I need help with my real analysis homework assignments questions 1,2 and 3. I've attached both the assignment and the "book" we use in the files.

I would like to thank Emiel Lorist and Bas Nieraeth for their support in the preparation of these notes. All the figures have been created by Emiel Lorist. I would also like to thank the students of the course on real analysis for pointing out the typos in the manuscript.

In Section1and2we introduceσ-algebras and measures. The Lebesgue measure is constructed in Section3and is based on AppendixBon Carath´eodory’s theorem. Uniqueness questions are addressed in AppendixAon Dynkin’s monotone class theorem. The amount of books on measure theory is almost not measurable. The lecture notes are based on [1], [8], [16]and[17]. A very complete treatment of measure theory is given in the impressive works [5].

In Sections5,6,7we introduce the integration theory and the Lebesgue spacesL p. This theory is fundamental in modern (applied) mathematics. There are many excellent books which give more detailed treatments on the sub ject. See for instance [1], [4], [6], [15] for detailed treatments.

In Section8we give a brief introduction to the theory of Fourier series. More thorough treat- ments can be found in for example [9], [10], [13], [14]and[19]. A full course on Fourier Analysis is offered as 3rd elective course based on the lecture notes [10]. The theory of Fourier series will be used in the 2nd year bachelor course on Partial Differential Equations [7], but also in several other parts of Mathematical Physics and Numerical Analysis.

We end this brief introduction with a quote from a historical note of Zygmund [18]:

“The Lebesgue integral did not arise via the theory of Fourier series but was created through the necessities of measuring geometric figures. But once it was introduced, it had an enormous impact on Analysis through Fourier series”:

•The Riesz-Fischer theorems7.5,8.7, in its initial formulation primarily a theorem on Fourier series.

•The M. Riesz interpolation theorem.

•Structure of sets of measure 0.

•The theory of trigonometric series has become a workshop of new methods in analysis, a place where new methods are first discovered before they are generalized and applied in other contexts.

•The theory of Fourier series gave a fresh impulse to problems of the differentiability of functions (Sobolev spaces etc.). MEASURE AND INTEGRATION 3 1.σ-algebras For a setSwe writeP(S) for its power set.

Definition 1.1.Le tSbe a s e t . A co l l ec t i o nR⊆P(S)i s ca l l ed aringif (i)∅∈R; (ii)A, B∈R=⇒B\A∈R; (iii)n∈N,A 1,A 2,...,A n∈R=⇒ n j=1 Aj∈R.

Re m a r k1.2.

(1) An equivalent definition is obtained if one replaces (iii)byA, B∈R=⇒A∪B∈R.This follows by induction.

(2) IfRis a ring, then for allA, B∈Rone hasA∩B∈R. Indeed, this follows from the identity A∩B=A\(A\B).

Definition 1.3.Le tSbe a s e t . A f a m i l yA⊆P(S)is cal led aσ-algebra 1if (i)∅,S∈A; (ii)A∈A=⇒A c∈A; (iii)A 1,A 2,...,∈A=⇒ ∞ j=1 Aj∈A.

The setsA∈Aa re o f t e n ca l l ed t h emeasurable subsetsofS.

Re m a r k1.4.

(1) IfAis aσ-algebra, then for allA 1,A 2,...,∈Aone has ∞ j=1 Aj∈A. This follows from the identity ∞ j=1 Aj= ∞ j=1 Ac j c.

(2) Everyσ-algebra is a ring. This follows from the identityB\A=B∩A c.

Example1.5.LetSbe a set.

(a)A={S,∅}is the smallest possibleσ-algebra onS.

(b)A=P(S) is the largest possibleσ-algebra onS.

Example1.6.LetS={1,2,3}.

(a) LetA={∅,S,{1},{2,3}}.ThenAis aσ-algebra.

(b) LetB={∅,S,{1},{2,3},{1,2}}.ThenBis not aσ-algebra.

Example1.7.LetSbe a set.

(a) The setR={A ⊆S:Ais finite}is a ring.

(b) Let 2A={A⊆S:Ais countable orA cis countable}.ThenAis aσ-algebra (see Exercise 1.2). It is called the countable-cocountableσ-algebra and is useful for counterexamples.

We continue with a more serious example which plays a crucial role in later constructions.

Example1.8.

(a) LetS=R.LetI 1be the collection of all sets of the form (a, b]witha≤b. These intervals will be calledhalf-open intervals.ThenI 1is not a ring since for instance (0,3]\(1,2] = (0,1]∪(2,3] is not inI 1.

(b) LetS=R.LetF 1be the collection of sets which can be written as a finite union of half-open intervals (thus of the form (a, b]witha≤b). ThenF 1is not aσ-algebra for several reasons. 3 We c h e c k t h a tF 1is a ring. (i) follows from∅=(1,1]∈F 1. (iii) is clear since a finite union of a finite union of intervals of the form (a, b] is again a finite union. It remains to check (ii). 1In part of the literature aσ-algebra is also called aσ-field2Recall that a setA⊆Sis countable ifAis finite or there is a bijectionf:N→A.

3For instance ∞ n=1 (0,1− 1n]=(0,1) is not inF 1. 4 MEASURE AND INTEGRATION For this we first note that it is simple to check that forA, B∈F 1one hasA∩B∈F 1and by induction this extends to the intersections of finitely many sets. For two intervals (a, b]and (c, d]usingB\A=B∩A candR\(a, b]=(−∞,a]∪(b,∞) we find (c, d]\(a, b]=(c, d]∩(R\(a, b]) = (c, d]∩(−∞,a] ∪ (c, d]∩(b,∞) =(c, a∧d]∪(b∨c, d].

This is inF 1again. Now ifA= m i=1 (ai,bi]andB= n j=1 (cj,dj]areinF 1,then B\A= n j=1 (cj,dj]\A= n j=1m i=1 (cj,dj]\(a i,bi].

and by the previous observations this is inF 1again.

(c) LetS=R d.Fora, b∈R dwitha=(α 1,...,α d)andb=(β 1,...,β d)withα j≤β jfor j∈{1,...,k}the half-open rectangles are given by (a, b]=(α 1,β1]×...×(α d,βd].

LetF dbe the collection of sets which can be written as a finite unions of half-open rectangles.

ThenFdis a ring (see Exercise1.7).

The proof of the next result is Exercise1.3.

Proposition 1.9(Intersection ofσ-algebras).Suppose thatA iis aσ-algebra onSfor everyi∈I.

Then i∈I Aiis aσ-algebra.

Definition 1.10.Le tSbe a set and letF⊆P(S). We writeσ(F)for the smal lestσ-algebra which containsF.Thenσ(F)i s ca l l ed t h eσ-algebrageneratedbyF. More precisely: 4 σ(F)= {A:Ais aσ-algebra onSandF⊆A}.

Example1.11.LetS={1,2,3},F={{1,2}}andG={{2,3},{1,2}}.

(a)σ(F)={∅,S,{1,2},{3}}.

(b)σ(G)=P(S). Indeed,{2,3} c={1},{1,2} c={3}and{2,3}∩{1,2}={2}.Thusthe singletons{1},{2}and{3}are inσ(G). Therefore, the required result follows since we can form every subset ofSby taking suitable finite unions.

Example1.12.LetS=NandF={{n}:n∈N}.Thenσ(F)=P(N).

Definition 1.13.Le t(S, d)be a metric space. 5Le tB(S)be t h eσ-algebra generated by the open sets inS.Thus B(S)=σ{open sets inS}.

Theσ-algebraB(S)is cal led 6theBorelσ-algebraofS. The sets ofB(S)are cal led theBorel subsetsofS.

Example1.14.One of the most importantσ-algebras is the Borelσ-algebra ofRwhich is usually denoted byB(R). Later on we will show thatB(R) =P(R).

The following lemma will be useful in some of the exercises on the Borelσ-algebra ofRandR d.

Lemma 1.15(Lindel¨of ).Le tA⊆R d. Assume that for eachi∈I,O i⊆R dis open. IfA⊆ i∈I Oi, then there exists a countableJ⊆Isuch thatA⊆ i∈J Oi 4Note that Proposition1.9ensures thatσ(F) is indeed aσ-algebra. The intersection makes sure we obtain the smallest possible one 5More generally one could take any topological space here6Named after the French mathematician F´elix Borel 1871-1956 MEASURE AND INTEGRATION 5 Proof.Choose for eachx∈A,i x∈Iandr x>0 such thatB(x, r x)⊆O ix.Foreachx∈Acho ose a x∈Q dands x∈Q∩(0,∞)suchthatx∈B(a x,sx)⊆B(x, r x)⊆O ix.LetF={B(a x,sx):x∈ A}. Then clearlyA⊆ x∈A B(a x,sx). Moreover,Fhas at most countably many sets. Indeed, this follows from the fact that it is a subset of{B(q, r):q∈Q d,r∈Q∩(0,∞)}which is countable.

Therefore, we can writeF={B(a xn,sxn):n∈N}withx n∈Afor eachn∈N.

Now letJ={i xn∈I:n∈N}.ThenA⊆ i∈J Oi. Indeed, ifx∈A,thenx∈B(a x,sx)and cho osingn∈Nsuch thata xn=a xands xn=s xwe find thatx∈O ix n ⊆ i∈J Oi Exercises Exercise1.1.LetS=RandF={A⊆R:A⊆[0,1] orA c⊆[0,1]}.IsFaring?

Exercise ∗1.2.Prove that the collection in Example1.7(b) is aσ-algebra.Hint:Use the following (well-known) facts: The subset of a countable set is again countable; The countable union of countable sets is again countable.

Exercise1.3.

(a) Prove Proposition1.9.

(b) Give an example of twoσ-algebrasAandBonS={1,2,3}such thatA∪Bis not aσ-algebra.

Exercise ∗1.4.LetSbe a set and letF={{s}:s∈S}be the collection consisting of all sets which contain one element ofS. Show thatσ(F) coincides with the countable-cocountable σ-algebra of Example1.7.

Exercise1.5.Show thatN,Q,R\Q∈B(R). That isN,QandR\Qare Borel subsets ofR.

Exercise ∗1.6.Consider the following collectionB 0={(−∞,x):x∈R}) of subsets ofR.

(a) Show thatσ(B 0) contains all open intervals.

(b) Show that every open set inRcan be written as the union of countably many open intervals.

Hint:Use Lindel¨of ’s Lemma1.15.

(c) Conclude thatσ(B 0)=B(R).

Exercise ∗1.7.LetI d⊆F dbe the collection of half-open rectangles of Example1.8.Provethe following assertions:

(a) IfI, J∈I dthenI∩J∈I d.

(b) IfI, J∈I d,thenI\Jis the union of finitely manydisjointsets fromI d,andthusI\J∈F d.

Hint:Use induction on the dimensiond. Use Example1.8(b) ford=1.

(c) EachA∈F dcan be written as union of finitely manydisjointsets inI d.

Hint:Use induction onnto prove this for all sets of the formA= n k=1 IkwithI 1,...,I n∈I d.

(d)F dis a ring.

Exercise ∗∗ 1.8.Prove that aσ-algebra is either finite or uncountable. 7 Hint:Recall thatP(N) is uncountable. 7This shows thatσ-algebras are either easy finite sets or quite complicated 6 MEASURE AND INTEGRATION 2.Measures Definition 2.1.Le tSbe a set andR⊆P(S)a ring. Letμ:R→[0,∞]be a function with 8 μ(∅)=0 (i)μi s ca l l edadditiveif for al l disjoint 9A1,...,A n∈Rone has μ n j=1 Aj = n j=1 μ(A j).

(ii)μi s ca l l edσ-additiveif for each disjoint sequence(A n)n≥1 inRwhich satisfies ∞ n=1 An∈R it holds that μ ∞ j=1 Aj = ∞ j=1 μ(A j).

Re m a r k2.2.

(1) By an induction argument it suffices to considern= 2 in the definition of additive. (See Exercise2.1).

(2) Ifμisσ-additive, then it is additive as follows by takingA m =∅form≥n+1.

Definition 2.3.Le tSbe a set and letAbe aσ-algebra onS.

(i) The pair(S,A)i s ca l l ed ameasurable space.

(ii) A functionμ:A→[0,∞]which satisfiesμ(∅)=0and which isσ-additive onAi s ca l l ed a measure. In this case the triple(S,A,μ)i s ca l l ed ameasure space.

(iii) If additional ly to (ii)μ(S)<∞,thenμis cal led a finite measure. If moreover,μ(S)=1, thenμi s ca l l ed aprobability measureand(S,A,μ)i s ca l l ed aprobability space. 10 Example2.4 (Counting measure).LetS=NandA=P(N). We write #Afor the number of elements of a finite setA, and we set #A=∞ifAis infinite. Letμ:A→[0,∞]begivenby μ(A)=#A.Thenμis a measure. Oftenμis denoted byτand called thecounting measure.

Example2.5 (Dirac measure/Dirac’s delta function).LetS=R,A=P(R). Letx∈R.Let δ x:A→[0,∞]begivenbyδ x(A)=1ifx∈Aandδ x(A)=0ifx∈R\A.Thenδ xis a measure.

It is usually called theDirac measure 11 atx.

Example2.6.LetSbe a set and letμ:P(S)→[0,∞]begivenbyμ(∅)=0andμ(A)=1if A =∅.IfScontains at least two elements, thenμis not a measure.

Example2.7 (Length of an interval). 12 LetR=F 1as in Example1.8.Fora≤bletλ((a, b]) = b−a(length of the interval). IfA=(a 1,b1]∪...∪(a n,bn] is a union of disjoint sets such that a j≤b jforj=1,...,n,thenwelet 13 λ(A)= n j=1 bj−a j.Thenλis additive. Later we will see thatλisσ-additive onF 1and has an extension to a measure onσ(F 1)=B(R).

Theorem 2.8.Le tRbe a ring andμ:R→[0,∞]be additive. The fol lowing assertions hold:

(i) IfA, B∈RandA⊆B,thenμ(A)≤μ(B)(monotonicity).

(ii) IfA 1,A 2,...∈Rare disjoint and ∞ j=1 Aj∈R,then μ ∞ j=1 Aj ≥ ∞ j=1 μ(A j). 8This assumption will always be made.9Here we meanA i∩A j=∅ifi =j 10Measure theory is at the very heart of probability theory. See the third year elective course11Named after the English theoretical physicist Paul Dirac 1902-1984. The “delta function” is actually not a function, but can be interpreted as a generalized function or as measure.

12See Section3for a further construction of the so-called Lebesgue measure13One can check that this does not depend on the way we write the setA. See below Definition3.6. MEASURE AND INTEGRATION 7 (iii) IfA 1,A 2,...∈Rand ∞ j=1 Aj∈Randμisσ-additive onR,then μ ∞ j=1 Aj ≤ ∞ j=1 μ(A j)(σ-subadditivity).

Proof.(i): WriteB=A∪(B\A). Then μ(B)=μ(A∪(B\A)) =μ(A)+μ(B\A)≥μ(A).

(ii): Letn∈N.Then n j=1 Aj⊆ ∞ j=1 Ajand therefore by additivity ofμ n j=1 μ(A j)=μ n j=1 Aj (i)≤μ ∞ j=1 Aj .

The result follows by lettingn→∞.

(iii): Let (2.1)B 1=A 1,B 2=A 2\A 1,B 3=A 3\(A 1∪A 2),etc.

Then (B n)n≥1 is a disjoint sequence and ∞ j=1 Bj= ∞ j=1 Aj. Therefore, by theσ-additivity ofμ μ ∞ j=1 Aj =μ ∞ j=1 Bj = ∞ j=1 μ(B j)(i)≤ ∞ j=1 μ(A j).

Let (a n)n≥1 be a sequence of real numbers. We writea n↑aif (a n)n≥1 is an increasing sequence which converges toa. Similarly, we writea n↓aif it decreases and converges toa. This notation will now be extended to sets.

Definition 2.9.Le tSbe a set.

(i) A sequence(A n)n≥1 of subsets ofSw i l l be ca l l edincreasingifA n⊆A n+1 for al ln∈N.

In this case we writeA n↑A,whereA= ∞ n=1 An.

(ii) A sequence(A n)n≥1 of subsets ofSw i l l be ca l l eddecreasingifA n⊇A n+1 for al ln∈N.

In this case we writeA n↓A,whereA= ∞ n=1 An.

Theorem 2.10.Le t(S,A,μ)be a m ea s u re s pa ce a n d l e t(A n)n≥1 be a s eq u e n ce i nA.

(i) IfA n↑A,thenμ(A n)↑μ(A).

(ii) IfA n↓Aandμ(A 1)<∞,thenμ(A n)↓μ(A).

Proof.(i): Define (B n)n≥1 as in (2.1). Then (B n)n≥1 is a disjoint sequence and the following identities hold: ∞ j=1 Bj=Aand n j=1 Bj=A n. Therefore, theσ-additivity ofμgives μ(A)= ∞ j=1 μ(B j) = lim n→∞n j=1 μ(B j)= lim n→∞ μ(A n).

(ii): See Exercise2.4. The following result will help us to checkσ-additivity on rings. It will be used in the construction of the Lebesgue measure in Lemma3.9.

Lemma 2.11(Sufficient condition forσ-additivity).Le tRbe a ring and letμ:R→[0,∞]be such thatμ(∅)=0andμis additive. Suppose that for each sequence(A n)n≥1 withA n↓∅one hasμ(A n)→0.Thenμisσ-additive onR. 8 MEASURE AND INTEGRATION Proof.Let (B j)j≥1 be a disjoint sequence inRwithB:= ∞ j=1 Bj∈R. We need to show that (2.2)μ(B)= ∞ j=1 μ(B j).

LetA n= ∞ j=n Bj=B\(B 1∪...∪B n−1 ). ThenA n∈RandA n↓∅. Now the assumption yields μ(A n)→0. On the other hand μ(B)=μ(A n∪B 1∪B 2∪...∪B n−1 )=μ(A n)+ n−1 j=1 μ(B j).

Therefore, μ(B)− n−1 j=1 μ(B j) =μ(A n)→0and(2.2) follows. Exercises Exercise2.1.LetRbearingonasetS. Assumeμ:R→[0,∞] satisfiesμ(∅)=0and μ(A∪B)=μ(A)+μ(B) for all setsA, B∈RwithA∩B=∅. Show thatμis additive.

Exercise2.2.LetRbe a ring on a setS.Letμ:R→[0,∞] be additive.

(a) Prove that forA, B∈Rwithμ(A)<∞andA⊆Bone has μ(B\A)=μ(B)−μ(A).

(b) Prove that forA, B∈Rwith μ(A)<∞one has μ(A∪B)=μ(A)+μ(B)−μ(A∩B).

(c) Show that for anyn∈Nand any sets (A j)n j=1 inR, finite subadditivityμ(A 1∪...∪A n)≤μ(A 1)+...+μ(A n).

Exercise2.3.Let (a n)n≥1 be numbers in [0,∞). Setμ(∅) = 0 and defineμ:P(N)→[0,∞]by μ(A)= n∈A an.Provethatμis a measure onN.

Exercise ∗2.4.

(a) Prove Theorem2.10(ii).

(b) Give an example of a measure space (S,A,μ) and setsA n ∈Asuch thatA n ↓∅and μ(A n)=∞for alln∈N.

Hint:Use the counting measure.

Exercise ∗2.5.Let (S,A,μ) be a measure space. ForA 1,A 2,...⊆Sdefine lim sup n→∞ An= ∞ k=1∞ n=k An.

(a) Show thats∈lim sup n→∞ Anif and only if there are infinitely manyn∈Nsuch thats∈A n.

(b) AssumeA 1,A 2,...∈A. Show that lim sup n→∞ An∈A.

(c) AssumeA 1,A 2,...∈Asatisfy ∞ n=1 μ(A n)<∞. Show thatμ(lim sup n→∞ An)=0. 14 Exercise ∗2.6.LetAbe theσ-algebra from Example1.7(b) withS=R. Defineμ:A→[0,∞] byμ(A)=0 ifAis countable andμ(A)=1 ifA cis countable. Show thatμis a measure. 14This is called the Borel-Cantelli lemma in probability theory. MEASURE AND INTEGRATION 9 3.Construction of measures It is not a simple task to construct a measure. In this section we will construct the Lebesgue measure onR dof which we have previously shown it is an additive mapping on the ringF 1in Example2.7. ToextendittotheBorelσ-algebra we use a deep result of Carath´eodory. 15 His result basically says that it is enough to check that a measure isσ-additive on a ring generating the desiredσ-algebra. A detailed proof can be found in TheoremB.6in the appendix, but it will do no harm if one takes the result for granted. 16 Theorem 3.1(Carath´eodory’s extension theorem).Le tSbeasetandletR⊆P(S)be a ring.

Suppose thatμ:R→[0,∞]isσ-additive onRandμ(∅)=0.Thenμextends to a measure μon σ(R). More precisely, there exists a measure μon(S, σ(R))such that μ(A)=μ(A)for al lA∈R.

Re m a r k3.2.The measure μis often unique. 17 When there is no danger of confusion we will write μagain for the extension toσ(R). However, in general one has to be careful about uniqueness.

For instance if we defineμon the ringF 1byμ(A)=∞ifA∈F 1is nonempty, thenμhas at least two extensions: the counting measure onB(R) is an extension ofμ, but also the measure ν:B(R)→[0,∞]givenbyν(A)=∞ifA =∅is an extension ofμ.

We continue with a uniqueness result which will be proved in the appendix.

Definition 3.3(π-system).A col lectionE⊆P(S)is cal led aπ-systemif for al lA, B∈Eone hasA∩B∈E.

Example3.4.

(a) Every ring is aπ-system.

(b) LetS=R d. The half-open rectanglesI dare aπ-system.

The following result will be proved in PropositionA.7.

Proposition 3.5(Uniqueness).Le tμ 1andμ 2both be measures on measurable space(S,A).

Assume the fol lowing conditions:

(i)E⊆Ais aπ-system withσ(E)=A; (ii)μ 1(S)=μ 2(S)<∞andμ 1(E)=μ 2(E)for al lE∈E.

Thenμ 1=μ 2onA.

We continue with the construction of theLebesgue 18 measureλ. In Example1.8we intro- duced the half-open intervals (a, b]∈I 1witha≤b. Also recall thatF 1denotes the collection of all finite unions of half-open intervals. We have seen thatF 1is a ring. Moreover, in of course every set inF 1can be written as a finite union of disjoint half-open intervals.

Definition 3.6(on unions of half-open intervals).Fo rA∈F 1of the formA=(a 1,b1]∪...∪ (a n,bn]with disjoint((a j,bj])n j=1 inI 1defineλ:F 1→[0,∞]as the sum of the lengths:

λ(A)= n j=1 (bj−a j).

The above is well-defined. To see this assumeA=(c 1,d1]∪...(c m,dm] is another representation ofAas a union of disjoint intervals. LetI ij=(c i,di]∩(a j,bj]. Then eitherI ij is empty or a half open interval, m i=1 Iij=(a j,bj]and n j=1 Iij=(c j,dj]. 15Carath´eodory 1873-1950 was a Greek mathematician working in Analysis, but also on Thermodynamics.16The appendix is not part of the exam17For instance whenμis a finite measure. See PropositionA.7.18Henri Lebesgue 1875–1941 was a French mathematician well-known for his integration theory. See Section5. 10 MEASURE AND INTEGRATION From the definition and the disjointness of the (I ij)m,n i,j=1 we obtain n j=1 λ (a j,bj] = n j=1 λ m i=1 Iij = n j=1m i=1 λ(I ij)= m i=1n j=1 λ(I ij)= m i=1 λ n j=1 Iij = m i=1 λ (c i,di] , which proves the well-definedness. Alternatively, one can observe thatλ(A) coincides with the Riemann integral of 19 1A. Indeed, fix an intervalIsuch thatA⊆I. By linearity of the Riemann integral I1Adx= n j=1 I1(aj,bj]dx= n j=1 (bj−a j)=λ(A).

In Example1.8we introduced the half-open rectangles (a, b]∈I dwitha=(α 1,...,α d)and b=(β 1,...,β d)andα j≤β jforj∈{1,...,d}. Also recall thatF ddenotes the collection of all finite unions of half-open rectangles. By Exercise1.7(d),F dis a ring. Moreover, in Exercise 1.7(c) it was shown that every set inF dcan be written as a finite union of disjoint half-open rectangles.

Definition 3.7(on unions of half-open rectangles).ForahalfopenrectangleI=(a, b]∈I dwith a=(α 1,...,α d)andb=(β 1,...,β d)let itsvolumebe denoted by |I|= d j=1 (βj−α j).

Fo rA∈F dof the formA=I 1∪...∪I nwith disjoint(I j)n j=1 inI ddefineλ d:F d→[0,∞]by λ d(A)= n j=1 |Ij|.

This is well-defined since for a rectangleR⊆R dwithA⊆R, we again have λ d(A)= R1Adx, wherethelatterisad-dimensional Riemann integral.

Re m a r k3.8.

(1) When the dimension is fixed and there is no danger of confusion we will writeλforλ d.Itis common to write|A|forA∈F das well.

(2) ForA∈F d,λ(A)=λ d(A)=|A|equals the volume ofA.

Next we want to extendλ dtoσ(F d)=B(R d) (see Exercise3.1for this identity). To apply Theorem3.1, we first need to check theσ-additivity ofλonF d. This will be done via Lemma 2.11.

Lemma 3.9.The functionλ:F d→[0,∞]isσ-additive onF d.

Proof.By Lemma2.11it suffices to prove each sequence (A n)n≥1 inF dwithA n↓∅, satisfies μ(A n)→0. Fixε>0. We have to findN∈Nsuch thatλ(A n)<εfor alln≥N.

Step 1:For eachn∈Ncho ose aB n∈F dsuch that Bn⊆A nandλ(A n\B n)≤2 −n ε.20 Since Bn⊆A nalso ∞ n=1 Bn=∅. It follows that{( Bn)c:n∈N}is an open cover of the set A1which is compact by the Heine-Borel theorem. Therefore, there exists anNsuch that A1⊆ N n=1 (Bn)c.It follows that N n=1 Bn⊆A c 1. Since all for alln≥1, Bn⊆A 1,wemusthavethat N n=1 Bn=∅. 19Recall that1 A(x)=1 ifx∈Aand1 A=0 ifx∈A c. 20So we just choose a setB nwhich is slightly smaller thanA n. MEASURE AND INTEGRATION 11 Step 2:LetC n= n j=1 Bjforn≥1. For everyn≥1,A n\C n= n j=1 (A n\B j)⊆ n j=1 (A j\B j).

Therefore, using Theorem2.8(i)in(∗) and Exercise2.2(c)in(∗∗), we find λ(A n\C n)(∗)≤λ n j=1 (A j\B j) (∗∗)≤ n j=1 λ(A j\B j)≤ n j=1 2−j ε<ε.

SinceC n=∅for alln≥N, we can conclude thatλ(A n)=λ(A n\C n)<εfor everyn≥N. We can now deduce the main result of this section.

Theorem 3.10(Lebesgue measure).There exists a unique measureλon(R d,B(R d))such that for al l half-open rectanglesI, one hasλ(I)=|I|,where|I|is the volume ofI. Moreover, for al l h∈R dandA∈B(R d),λ(A+h)=λ(A). 21 In the aboveA+h:={x+h:x∈A}.

Proof. Step 1: Existence.In Lemma3.9we have shown thatλisσ-additive on the ringF d.

Therefore, by Theorem3.1λextends to a measure onσ(F d)=B(R d) (see Exercise3.1).

Step 2: Uniqueness.Letμbe another measure such thatμ(I)=|I|for half-open rectangles I∈I d. Fixn∈Nand letS n=(−n, n] d. Defineλ (n) andμ (n) onB(R d)by λ (n) (A)=λ(A∩S n)andμ (n) (A)=μ(A∩S n).

Thenλ (n) andμ (n) are measures andλ (n) (R d)=λ(S n)=|S n|and similarlyμ (n) (R d)=|S n|.

Sinceλ (n) andμ (n) coincide onI d, it follows from Example3.4(b) and Proposition3.5that λ (n) =μ (n) onB(R d). Therefore, for anyA∈B(R d), sinceA∩S n↑ATheorem2.10yields λ (n) (A)=λ(A∩S n)→λ(A)andμ (n) (A)=μ(A∩S n)→μ(A).

Thusλ(A)=μ(A).

Step 3: Translation invariance:Leth∈R d. We claim that for everyA∈B(R d) one has A+h∈B(R d). For this letA h={A∈B(R d):A+h∈B(R d)}. By definitionA h⊆B(R d).

One can check thatA his aσ-algebra. For each open setAone hasA+his open and hence A+h∈B(R d). Therefore,B(R d)=σ({open sets})⊆A h, and the claim follows.

Defineμ honB(R d)byμ h(A)=λ(A+h). Thenμ his a measure and for any half-open rectangleI,μ h(I)=|I+h|=|I|=λ(I). By the uniqueness of step 2, we findμ h(A)=λ(A)for allA∈B(R d) and this proved the result. Re m a r k3.11.From TheoremB.6one can actually see that for anyA∈B(R d), λ(A)=inf ∞ j=1 |Ij|:A⊆ ∞ j=1 Ij,where (I j)j≥1 are disjoint half-open rectangles , but we will not use this formula.

Exercises on the Lebesgue measure If the dimension is fixed we writeλinstead ofλ dfor simplicity.

Exercise ∗3.1.LetF dbe as in Example1.8. Show thatσ(F d)istheBorelσ-algebraB(R d).

Hint:Use the Lindel¨of Lemma1.15.

Exercise3.2.

(a) Show that any countable subsetA⊆R dis inB(R d).

Hint:First show that{x}∈B(R d) for everyx∈R d.

(b) Show that any countable subsetA⊆R dsatisfiesλ(A) = 0. In particular,λ(Q d)=0.

Hint:First show thatλ({x})=0. 21This is calledtranslation invariance. Uptoascalingfactorλis the only measure onB(R d) which satisfies this property. See Exercise3.6. 12 MEASURE AND INTEGRATION Exercise ∗3.3.Fo ra, b∈R dwitha=(α 1,...,α d)andb=(β 1,...,β d)withα j≤β jfor j∈{1,...,d}let (a, b)=(α 1,β1)×...×(α d,βd)and[a, b]=[α 1,β1]×...×[α d,βd] be the open and closed rectangle, respectively. Prove that λ (a, b) =λ [a, b] =λ (a, b] and thus all coincide with the volume of the rectangle.

Hint:Use Theorem2.10.

Exercise ∗3.4 (Uncountable sets can have measure zero).Show that the Cantor setC⊆[0,1] is inB(R) and satisfiesλ(C)=0.

Hint:What can you say aboutλ(C n)?

Exercise ∗3.5.Fo rA⊆Randt≥0lettA={tx:x∈A}. Show that for eachA∈B(R), λ(tA)=tλ(A).

Hint:Use the same method as in the proof of Theorem3.10.

Exercise ∗3.6.Letμ:B(R)→[0,∞] be a measure such that for allh∈RandA∈B(R), μ(A+h)=μ(A). Letc=μ((0,1]) and assumec∈(0,∞). Prove the following assertions: 22 (a) For eachx≥0andq∈N,μ (0,qx] =qμ (0,x] .

(b) For eachp, q∈N,μ (0, pq] =c pq.

(c) For eachx≥0,μ (0,x] =cx.

(d) For eacha≤b,μ (a, b] =c(b−a).

(e) For eachA∈B(R),μ(A)=cλ(A).

Exercise ∗∗ 3.7 (Lebesgue-Stieltjes 23 measure).Leta

Exercises on general measures Exercise ∗3.8.LetAbe aσ-algebra onSand letT⊆S. Define the restrictedσ-algebraA Ton Tby A T={A∩T:A∈A}.

(a) Show thatA Tis aσ-algebra.

(b) IfT∈A, show thatA T={A⊆T:A∈A}.

(c) Letμbe a measure on (S,A). IfT∈Ashow that the restriction ofμtoA T is a measure again.

(d) IfAis the Borelσ-algebra on metric space (S, d), thenA Tcoincides with the Borelσ-algebra on (T, d).

Exercise3.9 (Non-uniqueness of extensions I).LetS={1,2,3,4}and letF={{1,2},{1,3}}.

Defineμ:F→[0,∞]byμ({1,2})=μ({1,3})= 12. Find two different extensions ofμto σ(F)=P(S). Why does this not contradict Proposition3.5?

Exercise ∗3.10 (Non-uniqueness of extensions II).LetS=Nand letF= {n, n+1,...}:n∈N .

(a) Show thatFis aπ-system.

(b) Show thatσ(F)=P(N) (c) Letτbe the counting measure and letμ:P(N)→[0,∞] be defined byμ(A)=2τ(A). Show thatτ=μonF. Why does this not contradict PropositionA.7? 22From this exercise we see that up to a scaling factor,λis the only translation invariant measure onB(R). The case for dimensionsd≥2 holds as well and can be proved in a similar way.

23Thomas Stieltjes 1856-1894 was a Dutch mathematician working in Analysis. He has even worked in Delft. MEASURE AND INTEGRATION 13 4.Measurable functions One of the aims will be to integrate functionsf:S→Rwith respect to a measureμon a measurable space (S,A). A way to do this is to use discretization in thera n g e 24 off.Sowewould like to know the measure of for instance the setA y,ε ={s∈S:f(s)∈[y, y+ε]}. Knowing this for ally∈Rand allε>0 makes it possible estimate the “area” underf. Of course we do need thesetstobeinAto make this work. This is the motivation of the definition of measurability.

See Section5for details on integration.

The natural setting to introduce measurability of functions is a follows:

Definition 4.1.Le t(S,A)and(T,B)be two measurable spaces. A functionf:S→Ti s ca l l ed measurableif for eachB∈B, one has 25 f−1 (B)∈A.

Re m a r k4.2.

(1) The composition of two measurable function is again measurable (see Exercise4.1).

(2) Instead off −1 (B)or{s∈S:f(s)∈B}one sometimes writes{f∈B}for the same set.

(3) In probability theory measurable functions are calledrandom variables.

It suffices to check measurability on a generating collectionF⊆B:

Lemma 4.3.Le t(S,A)and(T,B)be t w o m ea s u ra b l e s pa ce s a n d l e tf:S→T.SupposeF⊆B is such thatσ(F)=B.Iff −1 (F)∈Afor al lF∈F,thenfis measurable.

Proof.Define B={B∈B:f −1 (B)∈A}. Our aim is to show that B=B. We claim that B is aσ-algebra. Indeed, sincef−1 (∅)=∅∈Aalso∅∈ B. Similarly, sincef −1 (T)=S∈A, we findT∈ B.IfB∈ B,thenf −1 (T\B)=S\f −1 (B)∈A, which implies thatB c∈ B.If B 1,B 2,...∈ B,then f −1 ∞ n=1 Bn = ∞ n=1 f−1 (B n)∈A.

Therefore, the claim follows.

Now sinceF⊆ B, the claim yieldsB=σ(F)⊆ B⊆B. This implies B=B. In the sequel a metric spaceXwill always be equipped with its Borelσ-algebraB(X) (unless otherwise stated).

Proposition 4.4(Continuous mappings are measurable).Le t(X, d)and(Y, ρ)be metric spaces.

Iff:X→Yis continuous, thenfis measurable. 26 Proof.By the continuity offwe find that for all openO⊆Ythe inverse image isf −1 (O)open inXand hence inB(X). Since the open sets ofYgenerate the Borelσ-algebra, the result follows from Lemma4.3withF={O⊆Y:Ois open}. The most frequent case we will encounter is whenf:S→Rand (S,A) is a measurable space.

Unless otherwise stated we consider the Borelσ-algebraB(R)onR. The following characterization of measurability will be useful.

Proposition 4.5(Real valued functions).Le t(S,A)be a m ea s u ra b l e s pa ce . Fo rf:S→Rthe fol lowing are equivalent:

(i)fis measurable.

(ii) For al lr∈R, one hasf −1 ((−∞,r])∈A.

(iii) For al lr∈R, one hasf −1 ((−∞,r))∈A. 24In Riemann integration of functionsf:R d→Rthe discretization is always done in the domain of the function.

This is one of the ma jor differences with Lebesgue integration 25Recall thatf −1(B) is called the inverse image ofBbyfand is defined byf −1(B)={s∈S:f(s)∈B} 26The same result holds for topological spaces and the proof is the same. In the setting of Borel-σ-algebras, measurable functions are often called Borel measurable. 14 MEASURE AND INTEGRATION Proof.(i)⇒(ii) and (i)⇒(iii) are trivial.

(iii)⇒(i): LetF={(−∞,r):r∈R}. In Exercise1.6we have seen thatσ(F)=B(R).

Therefore, (i) follows from Lemma4.3.

(ii)⇒(i): This can be proved as before by proving the required version of Exercise1.6. Example4.6.LetS={1,2,3}andA={S,∅,{1,2},{3}}.Letf:S→Rbe given byf(s) = 2016 ifs= 1 andf(s)=0ifs =1. Thenfis not measurable, becausef −1 ({2016})={1}/ ∈A.Ifwe replaceAby (for example)A :={S,∅,{1},{2,3}},thenfbecomes measurable.

Recall thatx∨y=max{x, y}andx∧y=min{x, y}.

Theorem 4.7.Le t(S,A)be a m ea s u ra b l e s pa ce . Le tf, g:S→Rbe measurable functions and letα∈R. Then the fol lowing functions are al l measurable as wel l:

f+g, f−g, f·g, f∨g, f∧g, f +:=f∨0,f −:= (−f)∨0,|f|,α·f,1 f(iff =0onS).

Proof.We first claim thath:S→R 2given byh(s)=(f(s),g(s)) is measurable. To prove this observe that for all half-open rectanglesI=I 1×I 2⊆R 2, one has h −1 (I)={s∈S:f(s)∈I 1 andg(s)∈I 2}=f −1 (I1)∩g −1 (I2)∈A.

By Exercise3.1σ(half open rectangles) =σ(F 2)=B(R 2), we can use Lemma4.3to find thath is measurable.

To prove the statements we use that continuous functions are measurable (see Proposition4.4) and the fact that the composition of measurable functions is again measurable (see Exercise4.1).

For instance letϕ:R 2→Rbe given byϕ(x, y)=x+y.Thenϕis continuous and therefore measurable. Now writingf+g=ϕ◦hthe required measurability follows.

The proofs for the difference, product, maximum, minimum are similar. Note that the maximum x∨y= 12(x+y)+ 12|x−y|and this is a continuous function fromR 2toR. For the minimum there is an analogue formula.

The measurability off ±,|f|,α·fand 1fallfollowinthesamewaybyrewritingthemasφ◦f for a suitable continuous functionφ.

The case 1frequires some comment. Letφ:R\{0}→Rbe given byφ(x)= 1x. Note that in this caseR\{0}is a metric space on whichφis continuous, and therefore measurable by Proposition 4.4. Now for all open setsB∈B(R),C:=φ −1 (B)isopeninR\{0}and hence inB(R). Thus (φ◦f) −1 (B)=f −1 (C)∈A. Therefore, the measurability ofφ◦ffollows from Lemma4.3. Let R=[−∞,∞]. It will be useful to introduce measurability of functionsf:S→ Ras well.

For this, we introduce an analogue of the Borelσ-algebra on R.

Definition 4.8.Le tB( R)be t h eσ-algebra generated by the sets{∞},{−∞}andB∈B(R).The σ-algebraB( R)w i l l be ca l l ed t h e B o re lσ-algebra of R.

From Lemma4.3we see that a functionf:S→ Ris measurable if and only if{s∈S:f(s)= ±∞} ∈ Aandf −1 (B)∈Afor eachB∈B(R). We extend addition, multiplication, etc. to Rin the following way:

∞+a=a+∞=∞,for alla∈(−∞,∞] −∞+a=a−∞=−∞,for alla∈[−∞,∞) ∞·a=a·∞=∞,for alla∈(0,∞] ∞·a=a·∞=−∞,for alla∈[−∞,0) ∞·0=0·∞=a ∞=a −∞=0 for alla∈(−∞,∞) In this setting Theorem4.7remains true 27 for functionsf, g:S→ R. 27We do not define∞−∞, so some cases need to be excluded. For the proof one additionally needs to check in each of the cases that inverse images of{∞},{−∞},{0}are measurable. We leave this to the reader. MEASURE AND INTEGRATION 15 The next result shows that measurability is preserved under taking countable suprema, count- able infimum, limits of sequences, etc. For a sequence of numbers (x n)n≥1 in R,let lim sup n→∞ xn= lim k→∞ supn≥k xn and lim inf n→∞ xn= lim k→∞ infn≥k xn.

The limit ofy k:= sup n≥k xnexists 28 since (y k)k≥1 is decreasing. Moreover, (4.1) lim sup n→∞ xn=lim k→∞ yk=inf k≥1 yk=inf k≥1 supn≥k xn.

Similar formulas hold for the lim inf n→∞ xn. Recall that the lim sup n→∞ xnand lim inf n→∞ xn always exist. Moreover they both coincide with lim n→∞ xnif and only if (x n)n≥1 converges in R.

Theorem 4.9.Le t(S,A)be a m ea s u ra b l e s pa ce . Fo r ea c hn∈Nletf n:S→ Rbe a m ea s u ra b l e function. Then each of the fol lowing functions is measurable as wel l: 29 supn≥1 fn,inf n≥1 fn,lim sup n→∞ fn,lim inf n→∞ fn.

Moreover, iff n→fpointwise 30,thenfis measurable again.

Proof.Letg=sup n≥1 fn.Thenforeachr∈R, g −1 ([−∞,r]) ={s∈S:g(s)≤r}={s∈S:f n(s)≤rfor alln∈N}= ∞ n=1 f−1 n ([−∞,r])∈A.

Sinceσ({[−∞,r]:r∈R})=B( R), the measurability ofgfollows from Lemma4.3. The case of infima follows from inf n≥1 fn=−sup n≥1 (−f n).

By (4.1) we can write lim sup n→∞ fn=inf k≥1 sup n≥k fk. Therefore, the measurability follows from the previous cases. The remaining cases follow from lim inf n→∞ fn=−lim sup n→∞ (−f n) and lim n→∞ fn= lim sup n→∞ fn. Definition 4.10.A functionf:S→Ris cal led asimple function 31 iffis measurable and takes only finitely many values.

Lettingx 1,...,x n∈Rdenote thedistinctva l u e s o ffandA j={s∈S:f(s)=x j},wecan always represent a simple function as f= n j=1 xj·1 Aj.

Of course ifx j= 0 for somej∈{1,...,n},wecouldleaveitoutfromthesum.

Example4.11.LetS=RandA=B(R). The following functions are simple functions:

(a)f=π·1 (0,1) −4·1 (13,14] +5·1 Z∩(−∞,0) .

(b)f=1 Q.

Next we show that measurable functions can be written as limits of simple functions. For this we discretize in the range space in a suitable way. The result plays a crucial role in the integration theory in Section5.

Theorem 4.12.Le t(S,A)be a m ea s u ra b l e s pa ce . 32 (i) Letf:S→[0,∞]be measurable. Then there exists a sequence of simple functions(f n)n≥1 such that0≤f 1(s)≤f 2(s)≤...andlim n→∞ fn(s)=f(s)for eachs∈S.

(ii) Letf:S→ Rbe measurable. Then there exists a sequence of simple functions(f n)n≥1 such thatlim n→∞ fn(s)=f(s)for al ls∈S. 28Here we allow divergence to±∞29Here it is important what we work with countable suprema, infimum, etc.30Here we allow divergence to±∞31In part of the literature this is called a step function, but we will use this name for a different class of functions32In the following we allow divergence to±∞ 16 MEASURE AND INTEGRATION Proof. 33 (i): For eachn∈Nandj∈{0,1,...,4 n−1},let A n,j ={s∈S:j2 −n ≤f(s)<(j+1)2 −n }andA n={s∈S:f(s)≥2 n}.

Then for eachn, jone hasA n,j ,A n∈A. Define 34 fn=2 n1An+ 4n−1 j=0 j 2n1An,j .

It is clear that eachf ntakes finitely many values. Moreover, by Exercise4.5and Theorem4.7, eachf nis measurable. Thus eachf nis a simple function.

Now fixs∈S. We first prove 0≤f n(s)≤f n+1 (s)foreachn∈N. First assumef(s)<2 n.

Then selecting the uniquej∈{0,...,4 n−1}such thats∈A n,j we find thatf n(s)=j2 −n .

Similarly, we can selectk∈{0,...,4 n+1 −1}such thats∈A n+1,k and we find thatf n+1 (s)= k2 −(n+1) . Since,f(s)≥j2 −n =2j2 −(n+1) , we find thatk≥2jand therefore f n(s)=j2 −n ≤k2 −(n+1) =f n+1 (s).

The casef(s)≥2 ncan be treated similarly and is left to the reader.

To prove thatf n(s)→f(s), first assumef(s)<∞.Letε>0. ChooseN∈Nso large that f(s)<2 N and 2 −N <ε.Letn≥N. Selectingj∈{0,...,4 n−1}such thats∈A n,j we find that |f(s)−f n(s)|≤2 −n ≤2 −N <ε.

Therefore,f n(s)→f(s) in this case. Iff(s)=∞,thenf n(s)=2 nfor everyn∈Nand thus f n(s)→∞=f(s).

(ii): Writef=f +−f −. Then by (i) we can find simple functionsf n,+ ,fn,− :S→Rsuch thatf n,+ →f +andf n,− →f −.Letf n=f n,+ −f n,− forn∈N. Define the setsA +andA −by A ±={s∈S:±f(s)∈[0,∞]}.ThenA +∪A −=S.Ifs∈A +,then f n(s)=f n,+ (s)→f +(s)=f(s).

The cases∈A −is similar.

Exercises Exercise4.1.Let (S j,A j)forj=1,2,3 be measurable spaces. Assumef:S 1→S 2and g:S 2→S 3are both measurable. Show that the compositiong◦f:S 1→S 3is measurable.

Exercise4.2.Let (S,A) be a measurable space. Letf, g:S→Rbe measurable functions. Show that{s∈S:f(s)=g(s)}∈Aand{s∈S:f(s)

Exercise4.3.Let (S,A) be a measurable space. Letf:S→Rbe a measurable function and p∈(0,∞). Show that the function|f| pis measurable.

Exercise4.4.LetSbe a set. For a functionf:S→RletA f={f −1 (B):B∈B(R)}. Show thatA fis aσ-algebra. 35 Exercise4.5.Let (S,A) be a measurable space and letA⊆S. Show thatA∈Aif and only if the function1 A:S→Ris measurable.

Exercise4.6.Let (S,A) be a measurable space. Letf 1,f2,...:S→Rbe measurable functions andA 1,A 2,...∈Abe disjoint. Show that the functionf= ∞ n=1 1Anfnis measurable.

Exercise ∗4.7 (Vector-valued functions).Let (S,A) be a measurable space. For eachj∈{1,...,d} letf j:S→Rbe a function and letf:S→R dbe given byf=(f 1,...,f d). Prove thatfis measurable if and only iff jis measurable for eachj∈{1,...,d}.

Hint:For the “if part” one can use the same technique as in Theorem4.7. 33For the proof make a picture where you make a partition into intervals of length 2 −n on they-axis. For the picture put the setSis on thex-axis 34The idea is that for eachn∈Nwe approximatefup to 2 −n on the set{f<2 n}. 35Theσ-algebraA fis called theσ-algebra generated byf. It is the smallestσ-algebra for whichfis measurable. MEASURE AND INTEGRATION 17 Exercise ∗4.8 (Monotone functions).Assume thatf:R→Ris increasing. Show thatfis measurable, where as usually onRwe consider the Borelσ-algebra.

Hint:Use Proposition4.5.

Exercise ∗4.9 (Set of convergence).Let (S,A) be a measurable space. Letf 1,f2,...:S→Rbe measurable functions. Define A={s∈S:(f n(s)) n≥1 is convergent}.

(a) Explain whyA={s∈S:(f n(s)) n≥1 is a Cauchy sequence}.

(b) Show that for eachk, n, m∈Nthe setA(k, n, m)={s∈S:|f n(s)−f m(s)|< 1k}is inA.

(c) Show thatA= ∞ k=1∞ N=1∞ m=N∞ n=N A(k, n, m).

Hint:s∈Aif and only if∀k∈N∃N∈N∀n∈N∀m∈N:|f n(s)−f m(s)|< 1kand connect the universal quantifier with an intersection and the existential quantifier with a union.

(d) Conclude thatA∈A. 18 MEASURE AND INTEGRATION 5.Construction of the integral In this section (S,A,μ) is a measure space. Our goal is to construct an integral for measurable functionsf:S→ R. Notation:

Sfdμor Sf(s)dμ(s).

The integral will be built in three steps:

(1) For simple functionsf:S→[0,∞); (2) For measurable functionsf:S→[0,∞]; (3) For (certain) measurable functionsf:S→ R.

The advantage of this setting is that it works for any measure space (S,Σ,μ). Moreover, in the special case of the Lebesgue measure it extends the Riemann integral.

It will be convenient to use the following terminology.

Definition 5.1.For measurable functionsf, g:S→ Rwe say thatf=galmost everywhere ifμ({s∈S:f(s) =g(s)})=0. Notation:f=ga.e.36 Similarly, one can definef<∞a.e., etc.

5.1.Integral for simple functions.

Definition 5.2(Integral for simple functions).Le tf:S→[0,∞]be a simple function given by (5.1)f= n j=1 xj·1 Aj, withx 1,...,x n∈[0,∞]and(A j)n j=1 disjoint sets inA.ForE∈Alet 37 Efdμ= n j=1 xj·μ(E∩A j).

This is cal led theintegraloffover the setE.

Re m a r k5.3.By using a common refinement one checks that if a different representation for the functionffrom (5.1) is used, then the integral overEgives the same value (see Lemma5.4(ii)).

Clearly Efdμ∈[0,∞] for everyE∈A.

We continue with some basic properties of the integral.

Lemma 5.4.Le tf, g:S→[0,∞]be simple functions. Then the fol lowing hold:

(i) For al lE∈A, Efdμ= S1Efdμ.

(ii) (monotonicity I) IfE∈Aandf≤gonE,then Efdμ≤ Egdμ.

(iii) (monotonicity II) IfE, F∈AsatisfyE⊆F,then Efdμ≤ Ffdμ.

(iv) (linearity) For al lE∈Aandα, β∈[0,∞), Eαf+βgdμ=α Efdμ+β Egdμ.

(v) (additivity) For al l disjoint setsE 1,E 2∈A, E1∪E 2fdμ= E1fdμ+ E2fdμ.

(vi) Sfdμ=0if and only iff=0almost everywhere. 38.

Proof.(i). This is immediate from the definition and the identity1 E∩A =1 E·1 A.

For the proof of the remaining assertions we continue with a preliminary observation. Write f= m i=1 xi·1 Aiwith (A i)m i=1 disjoint sets inAwith m i=1 Ai=S,andg= n j=1 yj·1 Bjwith 36In probability theory this is usually called almost surely and this is abbreviated as a.s.37Here we use the convention 0·∞=038See Definition5.1 MEASURE AND INTEGRATION 19 (B j)n j=1 disjoint sets inAwith m i=1 Bi=S.LetC i,j =A i∩B jfor alliandj.Then(C i,j)m,n i,j=1 are disjoint sets inA. By additivity Efdμ= m i=1 xi·μ(E∩A i)= m i=1n j=1 xi·μ(E∩C i,j) (5.2) Egdμ= n j=1 yj·μ(E∩B j)= n j=1m i=1 yj·μ(E∩C i,j). (5.3) (ii): For disjoint setsA, B⊆Sone has1 A∪B =1 A+1 B and hence (5.4)1 Ef= m i=1n j=1 xi·1 E∩C i,j and1 Eg= n j=1m i=1 yj·1 E∩C i,j.

Therefore,x i≤y jwheneveri, jsatisfyE∩C i,j =∅. This together with (5.2), (5.3), yields (ii).

(iii): This follows fromμ(E∩A i)≤μ(F∩A i) which is immediate from the monotonicity ofμ.

(iv): By (5.4)withE=S, we can writeαf+βg= m i=1 n j=1 (αx i+βy j)·1 Ci,j. Therefore, Eαf+βgdμ= m i=1n j=1 (αx i+βy j)μ(E∩C i,j) =α m i=1n j=1 xiμ(E∩C i,j)+β n j=1m i=1 yjμ(E∩C i,j)=α Efdμ+β Egdμ, where we used (5.2)and(5.3).

(v): Since1 E1∪E 2f=1 E1f+1 E2f, E1∪E 2fdμ (i)= S1E1∪E 2fdμ (iv)= S1E1fdμ+ S1E2fdμ (i)= E1fdμ+ E2fdμ.

(vi): By leaving out some of thosekfor whichx k= 0, we can assumex i>0 for alli.Let A={s∈S:f(s)>0}and observe thatA= n i=1 Ai.Now Sfdμ= m i=1 xiμ(A i)with μ(A i)≥0foreachi. Therefore, if the integral is zero, thenμ(A i)=0foreachiand hence μ(A)= n i=1 μ(A i) = 0. Conversely, iff= 0 a.e., monotonicity yieldsμ(A i)≤μ(A) = 0, and the result follows. Example5.5.Letλbe the Lebesgue measure onR.SinceQ∈B(R) it follows from Exercises1.5 and4.5that1 Qis measurable and by Exercise3.2for anyE∈B(R), E1Qdλ=λ(E∩Q)=0.

Note that1 Qis not Riemann integrable on any interval [a, b]witha

5.2.Integral for positive measurable functions.In order to extend the definition of the integral to arbitrary measurable functionsf:S→[0,∞] we use the following lemma. In the sequel, we writef n↑fif for alls∈S,(f n(s)) n≥1 is increasing andf n(s)→f(s).

Lemma 5.6(Consistency).Le tf:S→[0,∞]be a measurable function. Suppose that(f n)n≥1 is sequence of simple functions such that0≤f n↑f.Supposeg:S→[0,∞]is a simple function such that0≤g≤f.Then Egdμ≤lim n→∞ Efndμfor everyE∈A.

Observe that lim n→∞ Efndμexists in [0,∞], because it is an increasing sequence of real numbers.

Proof.Letε∈(0,1) and setE n={s∈E:(1−ε)g(s)≤f n(s)}forn∈N. Then using the indicated parts of Lemma5.4in the estimates below, we find (1−ε) Engdμ (iv)= En(1−ε)gdμ (ii)≤ Enfndμ (iii)≤ Efndμ (ii)≤lim n→∞ Efndμ. 20 MEASURE AND INTEGRATION It remains to prove Engdμ→ Egdμ. For this writeg= m i=1 xi·1Aiwith (A i)m i=1 inAdisjoint andx 1,...,x m ≥0. SinceE n∩A i↑E∩A iasn→∞,Theorem2.10yields (5.5) Engdμ= m i=1 xi·μ(E n∩A i)→ m i=1 xi·μ(E∩A i)= Egdμ.

Definition 5.7(Integral for positive functions).For a measurable functionf:S→[0,∞]and a sequence of simple functions(f n)n≥1 with0≤f n↑fwe define theintegraloffoverE∈Aas Efdμ= lim n→∞ Efndμin[0,∞].

The above limit exists in [0,∞]sincethenumbersa n:= Efndμform an increasing sequence (a n)n≥1 in [0,∞]. Note that by Theorem4.12we can always find simple functionsf n:S→[0,∞) such that 0≤f n↑f. However, we need to check that the above definition does not depend on the choice of the sequence (f n)n≥1 .Let(g m)m≥1 be another sequence of simple functions such that 0≤g m ↑fand letb m = Egmdμ. It suffices to show that lim n→∞ an= lim m→∞ bm. By Lemma 5.6for eachm≥1, b m = Egmdμ≤lim n→∞ Efndμ= lim n→∞ an.

From this we obtain lim m→∞ bm ≤lim n→∞ an. Reversing the roles ofg m andf n, one sees that the converse holds as well.

Next we can extend the properties of the integral of Lemma5.4.

Proposition 5.8.Le tf, g:S→[0,∞]be measurable functions. Then the fol lowing hold:

(i) For al lE∈A, Efdμ= S1Efdμ.

(ii) (monotonicity I) IfE∈Aandf≤gonE,then Efdμ≤ Egdμ.

(iii) (monotonicity II) IfE, F∈AsatisfyE⊆F,then Efdμ≤ Ffdμ.

(iv) (linearity) For al lE∈Aandα, β∈[0,∞), Eαf+βgdμ=α Efdμ+β Egdμ.

(v) (additivity) For al l disjoint setsE 1,E 2∈A, E1∪E 2fdμ= E1fdμ+ E2fdμ.

(vi) Sfdμ=0if and only iff=0almost everywhere.

Proof.(i): Let (f n)n≥1 be simple functions such that 0≤f n↑f. Then by Lemma5.4(i), Efdμ= lim n→∞ Efndμ= lim n→∞ S1Efndμ= S1Efdμ.

(ii)–(v): See Exercise5.2.

(vi): Assumef= 0 a.e. Choose a sequence of simple functions (f n)n≥1 such that 0≤f n↑f.

Thenf n= 0 a.e. for eachn∈Nand thus by Lemma5.4(vi), Sfdμ= lim n→∞ Sfndμ=0.

For the converse we use contraposition. Assume one does not havef= 0 a.e. ThenE={s∈ S:f(s)>0}satisfiesμ(E)>0. LettingE n={s∈S:f(s)≥ 1n}we findE n↑E,sobyTheorem 2.10,μ(E n)→μ(E). Therefore, there exists ann∈Nsuch thatμ(E n)>0 and thus Sfdμ (iii)≥ Enfdμ (ii)≥ En 1ndμ= 1nμ(E n)>0.

Example5.9 (Series are integrals).LetS=NandA=P(N). Letτ:P(N)→[0,∞]denotethe counting measure. Letf:N→[0,∞] be arbitrary. Thenfis clearly measurable. Now define for eachn≥1,f n= n j=1 f(j)1 {j} . Since eachf nis a simple function and 0≤f n↑fwe find Nfdτ= lim n→∞ Nfndτ= lim n→∞n j=1 f(j)= ∞ j=1 f(j). MEASURE AND INTEGRATION 21 5.3.Integral for measurable functions.The final step is to define the integral for measurable functionsf:S→ Rby using the splittingf=f +−f −. Recall thatf + =max{f,0}and f −=max{−f,0}. Note that|f|=f ++f −.

Definition 5.10.A measurable functionf:S→ Ri s ca l l edintegrablewhen both Sf+dμand Sf−dμare finite. In this case the integral offoverE∈Ais defined as Efdμ= Ef+dμ− Ef−dμ.

Re m a r k5.11.

(1) Iff:S→ Ris integrable, then by monotonicity Ef+dμ∈[0,∞)and Ef−dμare finite for everyE∈A. Therefore, Efdμis a number inR.

(2) A measurable functionf:S→[0,∞] is integrable if and only if Sfdμ<∞.

(3) Iff:S→ Ris integrable, thenμ({s∈S:f(s)=±∞}}= 0 (see Exercise5.1).

To check integrability the following characterization is very useful.

Proposition 5.12.For a measurable functionf:S→ Rthe fol lowing are equivalent:

(i)fis integrable; (ii)|f|is integrable.

Moreover, in this case for eachE∈A, (triangle inequality) Efdμ ≤ E|f|dμ.

Proof.First observe that|f|=f ++f −. Therefore, both assertions are equivalent toI +,I−<∞, whereI ±= Sf±dμand hence the result follows. To prove the required estimate note that by the triangle inequality and linearity of the integral for positive functions:

Efdμ =|I +−I −|≤I ++I −= Ef++f −dμ= E|f|dμ.

By considering the positive and negative part separately one can check Proposition5.8(i),(ii), (v) hold for all integrable functionsf, g:S→ Ragain (see Exercise5.4). The following example shows that Parts (iii)and(vi) do not extend to this setting.

Example5.13.LetS=Randλthe Lebesgue measure. Letf=1 (0,1] −1 (1,2] .Thenf +=1 (0,1] andf −=1 (1,2] and (0,1] fdλ= (0,1] f+dλ=λ((0,1]) = 1, (0,2] fdλ= (0,2] f+dλ− (0,2] f−dλ=λ((0,1])−λ((1,2]) = 1−1=0.

The extension of the linearity (iv) is more difficult and proved below.

Proposition 5.14.Le tf, g:S→ Rbe integrable functions andα, β∈R.Then 39 αf+βgis integrable, and for al lE∈A (linearity) Eαf+βgdμ=α Efdμ+β Egdμ.

Proof.Note that by Proposition5.12each of the following functions is integrablef ±,g ±,|f|,|g|.

Therefore,|α||f|+|β||g|is integrable by Proposition5.8(iv). Since|αf+βg|≤|α||f|+|β||g|, the functionαf+βgis integrable as well (see Exercise5.7). 39Of course we only consider those functions for whichαf+β+gis well-defined. 22 MEASURE AND INTEGRATION It remains to prove the identity for eachE∈A. To do this we first consider the caseα=β=1.

Wr i t ef+g=φ−ψ,whereφ=f ++g +andψ=(f −+g −). Then by Proposition5.8(iv)φand ψare integrable and Exercise5.3yields (5.6) Ef+gdμ= Eφdμ− Eψdμ = Ef+dμ+ Eg+dμ− Ef−dμ− Eg−dμ= Efdμ+ Egdμ.

It remains to show that for allα∈R, Eαfdμ=α Efdμ.

Ifα≥0, thenαf=(αf) +−(αf) −=αf +−αf −.Proposition5.8(iv) yields Eαfdμ= Eαf +dμ− Eαf −dμ=α Ef+dμ−α Ef−dμ.

In caseα<0, then by (5.6) and the previous case, we find 0= Eαf+(−α)fdμ= Eαfdμ+ E(−α)fdμ= Eαfdμ+(−α) Efdμ and the result follows by subtracting the second term on both sides. Example5.15.Every simple functionf:S→Rgiven byf= n j=1 xj1AjwithA 1,...,A n∈A andμ(A j)<∞for allj, is integrable and by Proposition5.14, Efdμ= n j=1 xj E1Ajdμ= n j=1 xjμ(E∩A j).

In the next example we show that for continuous functions the integral with respect to the Lebesgue measure coincides with the Riemann integral. 40 Example5.16.Leta

Moreover, since there is a constantM≥0 such that|f|≤M,wecanuseExercise5.7to deduce thatfis (Lebesgue) integrable. Sincefis continuous, it is Riemann integrable. Below we show that L(f)=R(f), where we used the abbreviations L(f):= [a,b] fdλ(Lebesgue integral) and R(f):= b a f(x)dx(Riemann integral).

Toprovethisidentityletε>0. Sincefis uniformly continuous we can find aδ>0 such that for allx, y∈[a, b],|x−y|<δimplies|f(x)−f(y)|< εb−a .Letn∈Nbe such that b−an <δ.Let x j=a+j b−an forj=0,...,n.Letm j=min{f(x):x∈[x j−1 ,xj]}andM j=max{f(x):x∈ [x j−1 ,xj]}.Letg(x)=1 {a} f(a)+ n j=1 1(xj−1 ,xj]mjandG(x)=1 {a} f(a)+ n j=1 1(xj−1 ,xj]Mj.

ThengandGare integrable in the sense of Riemann and Lebesgue and one can check L(g)=R(g)= n j=1 (xj−x j−1 )m j=:αand L(G)=R(G)= n j=1 (xj−x j−1 )M j=:β.

Sinceg≤f≤G, by monotonicity we find that L(f),R(f)∈[α, β]. Therefore, |R(f)−L(f)|≤β−α= n j=1 (xj−x j−1 )(M j−m j)≤ n j=1 b−a nε b−a=ε, Sinceε>0 is arbitrary, we find that R(f)=L(f). 40It is even known that every Riemann integrable functionfis Lebesgue integrable and the integrals coincide (see [1, Theorem 23.6]). The proof uses a similar method as ours. However, one should be aware that not every improperRiemann integral is Lebesgue integrable. See Theorem6.12and Exercise6.10for more details on improper Riemann integrals MEASURE AND INTEGRATION 23 Example5.17.LetS=R,A=P(R). Letx∈Rand letδ xbe the Dirac measure from Example 2.5. Then for anyf:R→R.

Rfdδ x=f(x).

Indeed, for simple functionsfthis is obvious. Forf:S→[0,∞] this follows by approximation from below by simple functions. For generalf:S→[0,∞] this follows by writingf=f +−f −.

Example5.18. 41 Consider the setting of the counting measure of Example5.9. A functionf:

N→Ris integrable if and only if ∞ j=1 |f(j)|<∞. In that case Nfdτ= ∞ j=1 f(j).

Finally we briefly indicate how one can extend the integral to complex functionsf:S→C.

On the complex numbersCwe consider its Borelσ-algebra. Iff:S→Cis measurable, then we writef=u+ivwithu, v:S→Randuandvare measurable (see Exercise5.11).

Definition 5.19.A measurable functionf:S→Cgiven byf=u+ivwithu, v:S→Ris ca l l edintegrableif bothuandvare integrable. In this case let Efdμ= Eudμ+i Evdμ, E∈A.

Exercise5.11yields that Propositions5.12and5.14extend to the complex setting.

Exercises about general integrals In the exercises below (S,A,μ) denotes a measure space.

Exercise5.1.Letf:S→[0,∞] be an integrable function. Show thatf<∞a.e.

Exercise5.2.

(a) Prove Proposition5.8(iv).

Hint:Approximate by simple functions as in the definition of the integral.

(b) Prove Proposition5.8(ii), Hint:Wr i t e1 Eg=1 E(g−f)+1 Efand use Proposition5.8(iv).

(c) Prove Proposition5.8(iii).

(d) Prove Proposition5.8(v).

ThefollowingwasusedintheproofofProposition5.14.

Exercise5.3.Assumeg, h:S→[0,∞]andf:S→ Rare all integrable functions andf=g−h.

Show that Efdμ= Egdμ− Ehdμ, E∈A.

Hint:Note thatf ++h=f −+gand use Proposition5.8(iv).

Exercise5.4.Extend the following results to integrable functionsf, g:S→ R:

(a) Proposition5.8(i). First show that1 Efis integrable forE∈A.

(b) Proposition5.8(ii).

(c) Proposition5.8(v) Exercise5.5.Letf, g:S→ Rboth be measurable functions. Assumefis integrable andf=g a.e. Show thatgis integrable and Efdμ= Egdμ, E∈A.

Hint:First try to prove this in the casefandgtake values inR. In this case apply Proposition 5.8(vi)toh=|f−g|. For the case wherefandgare R-valued one needs a careful analysis of positive and negative parts. 41For details see Exercise5.8 24 MEASURE AND INTEGRATION Exercise5.6.Letλdenote the Lebesgue measure on (R,B(R)). Letf:R→Rbe an integrable function. Show that for all−∞

Is this true if we replaceλby an arbitrary measure onB(R)?

Exercise ∗5.7 (Domination).Letf, g:S→Rbe measurable functions.

(a) Assume|f|≤gandgis integrable. Prove thatfis integrable.

(b) Show thatfis integrable if and only if ∞ n=−∞ 2nμ({s∈S:2 n<|f(s)|≤2 n+1 })<∞.

Hint:Use (a) for a suitable functiong:S→[0,∞].

In the next exercise you are asked to give the details of Example5.18 Exercise ∗5.8.Consider the setting of the counting measure of Example5.9.

(a) Show that a functionf:N→Ris integrable if and only if ∞ n=1 |f(n)|<∞.

(b) Assume ∞ n=1 |f(n)|<∞. Show that Nfdτ= ∞ j=1 f(j).

Hint:Wr i t ef=f +−f −and approximatef +andf −in a similar way as in Example5.9.

Exercise ∗5.9 (Chebyshev’s inequality).Letf:S→[0,∞] be a measurable function. Prove that for everyt>0, μ({s∈S:f(s)≥t})≤1 t Sfdμ.

Hint:LetA t={s∈S:f(s)≥t}and writeμ(A t)= S1Atdμ.

Exercise ∗5.10.Letλbe the Lebesgue measure on (R d,B(R d)). Letf:R d→Rbe a measurable function. Forh∈R ddefine the translationf h:R d→Rbyf h(x)=f(x−h).

(a) Show thatf his measurable.

(b) Assume thatfis integrable. Show thatf his integrable and Rdfhdλ= Rdfdλ.

Exercise ∗∗ 5.11 (Complex functions).

(a) Letf:S→Cbe a measurable function and writef=u+ivwithu, v:S→R. Show thatu andvare measurable.

(b) Prove Proposition5.14for integrable functionsf, g:S→Candα, β∈C.

(c) Prove Proposition5.12for integrable functionsf:S→C.

(d) Which parts of Proposition5.8remain true for integrable functionsf, g:S→C? MEASURE AND INTEGRATION 25 6.Convergence theorems and applications In this section (S,A,μ) is a measure space.

One of the problems with the Riemann integral is that the cases where limit and integral can be interchanged are rather limited. In modern mathematics it is crucial to have better “tools” for this. We use the integration theory developed so far in order to obtain these tools. In Section 6.1we prove three famous convergence results. In Section6.2we give several consequences and applications.

We start with several simple examples which illustrate some difficulties.

Example6.1.For instance if (x j)j≥1 is an enumerate ofQ,andA n={x 1,...,x n},then1 An→1 Q pointwise. Each1 Anis Riemann integrable, but1 Qis not.

Example6.2.Letf n:R→Rbe defined by one of the followingn1 (0, 1n],1n1(n,2n) or1 (n,∞) for n≥1. Thenf n→0, but in each case Rfndλ →0.

6.1.The three main convergence results.In this section we prove the three most famous convergence results of integration theory:

•Monotone Convergence Theorem (MCT); 42 •Fatou’s lemma; 43 •Dominated Convergence Theorem (DCT). 44 Theorem 6.3(Monotone Convergence Theorem (MCT)).Le t(f n)n≥1 be measurable functions such that0≤f n↑f.Thenfis measurable and lim n→∞ Sfndμ= Sfdμ.

Proof.By Theorem4.9,f:S→[0,∞] is measurable. By monotonicity of the integral α:= lim n→∞ Sfndμ≤ Sfdμ.

It remains to prove Sfdμ≤α. For this choose a sequence of simple functions (g m)m≥1 such that 0≤g m ↑f. We claim that Sgmdμ≤αfor eachm≥1. The required estimate follows from the claim by lettingm→∞and using the definition of Sfdμ. To prove the claim we repeat part of the argument of Lemma5.6. Fixm∈Nand writeg=g m.

Letε∈(0,1) and for eachn≥1, setE n={s∈S:(1−ε)g(s)≤f n(s)}. Then using the indicated parts of Proposition5.8in the estimates below, we find (1−ε) Engdμ (iv)= En(1−ε)gdμ (ii)≤ Enfndμ (iii)≤ Sfndμ (ii)≤α.

Finally, Engdμ→ Sgdμfollows from (5.5)withE=Sin Lemma5.6. Therefore, we can conclude (1−ε) Sgdμ≤αand the claim follows sinceε∈(0,1) was arbitrary. Lemma 6.4(Fatou’s Lemma).Le t(f n)n≥1 be a sequence of measurable functions with values in [0,∞].Then Slim infn→∞ fndμ≤lim inf n→∞ Sfndμ.

Proof.Note thatf= lim inf n→∞ fnis a measurable function by Theorem4.9. Fixn∈Nand let g n=inf k≥n fk. For allm≥n,wehaveg n≤f m and by the monotonicity of the integral this gives Sgndμ≤ Sfmdμ. Therefore, taking the infimum over allm≥n, we find (6.1) Sgndμ≤inf m≥n Sfmdμ. 42This result is due to Beppo Levi 1875–1961 who was an Italian mathematician.43This result is named after the French mathematician Pierre Fatou 1878–1929.44This is due to Lebesgue (see footnote18) 26 MEASURE AND INTEGRATION Note that 0≤g n↑fand thus Sfdμ= S limn→∞ gndμ (MCT)= lim n→∞ Sgndμ (6.1)≤lim n→∞ infm≥n Sfmdμ= lim inf n→∞ Sfndμ.

Theorem 6.5(Dominated Convergence Theorem (DCT)).Le t(f n)n≥1 be a s eq u e n ce o f m ea s u r - able functions such thatlim n→∞ fn=fpointwise. If there exists an integrable functiong:S→[0,∞] such that|f n|≤gfor al ln≥1,thenf nandfare integrable and (6.2) lim n→∞ Sfndμ= Sfdμ.

Proof.The functionfis measurable by Theorem4.9. Moreover, since also|f|≤g,weseethat fis integrable. The same holds for eachf n.Letx n:= Sfndμandx:= Sfdμ. It suffices to show that lim sup n→∞ xn≤x≤lim inf n→∞ xn. It follows that Sgdμ±x= Sg±fdμ= S limn→∞ (g±f n)dμ(linearity) ≤lim inf n→∞ Sg±f ndμ(Fatou’s lemma withg±f n≥0) = lim inf n→∞ Sgdμ±x n (linearity).

= Sgdμ+ lim inf n→∞ (±x n).

Since Sgdμ<∞, we find that±x≤lim inf n→∞ (±x n). This implies the estimates x≤lim inf n→∞ xn and lim sup n→∞ xn≤x, where for the second part we used lim inf n→∞ (−x n)=−lim sup n→∞ xn. For functionsf, f 1,f2:S→ Rwe say thatf n→fa.e. if there exists a setA∈Asuch that μ(A) = 0 and for alls∈S\A,f n(s)→f(s). For the details of the following remark we refer to Exercise6.5.

Re m a r k6.6.Letf, f 1,f2,...:S→ Rbe measurable functions.

(1) Theorem6.3also holds under the weaker assumption that 0≤f n↑fa.e.

(2) Theorem6.5also holds under the weaker assumptions thatf n→fa.e. and|f n|≤ga.e.

Example6.7.Letf:R→ Rbe integrable with respect to the Lebesgue measureλ.LetF:R→R be given byF(x)= (−∞,x] fdλ.ThenFis continuous onR. Indeed, ifx n

6.2.Consequences and applications.We continue with several consequences and applications.

We start with a result on integration of series of positive functions.

Corollary 6.8(Series and integrals).Le tf 1,f2,...:S→[0,∞]be measurable functions. Then (6.3) S∞ n=1 fndμ= ∞ n=1 Sfndμ.

Proof.This follows from the MCT (See Exercise6.4). From a measureμone can build many other measures in the following way: MEASURE AND INTEGRATION 27 Theorem 6.9(Density).Le tf:S→[0,∞]be a measurable function. Defineν:A→[0,∞]by ν(A)= Afdμ.

Thenνis a measure. 45 Moreover,g:S→ Ris integrable with respect toνif and only iffgis integrable with respect toμ. In this case (6.4) Sgdν= Sfgdμ Proof. Step 1:Clearly,ν(∅)=0. For (A n)n≥1 a disjoint sequence inA,andA:= ∞ n=1 An, ν(A)= S1Afdμ= S∞ n=1 1Anfdμ (6.3)= ∞ n=1 S1Anfdμ= ∞ n=1 ν(A n).

Step 2:First we show that (6.4) holds for all measurableg:S→[0,∞]. Forg=1 A this is immediate from S1Adν=ν(A)= S1Afdμ. For simple functionsg:S→[0,∞) this follows by linearity. For a measurable functiong:S→[0,∞], by Theorem4.12we can find a sequence of simple functions (g n)n≥1 such that 0≤g n↑g. By the previous case, we obtain Sgdν (MCT)= lim n→∞ Sgndν= lim n→∞ Sfg ndμ (MCT)= Sfgdμ.

Step 3:To prove the “if and only if ” assertion and (6.4), letg:S→ Rbe a measurable function. Since step 2 yields that Sg±dν= Sfg ±dμ, both the equivalence and (6.4) follows by writingg=g +−g −. Example6.10.Letf:R→[0,∞] be measurable. Defineν:B(R d)→[0,∞)by ν(A)= Afdλ, A∈B(R).

Thenνis a measure. For example one could takef(x)= 1√2πe−x22.Thenνis the standard Gaussian measure onR.

In Example5.16we have discussed a connection between the Riemann and Lebesgue integral.

A similar connection holds for improper Riemann integrals.

Definition 6.11.Le tf:R→Rbe a continuous function. We say that ∞ −∞ f(x)dxexists as an improperRiemann integral if the limitsL 1= lim t→∞ t 0f(x)dxandL 2= lim t→−∞ 0 tf(x)dx exist inR. In that case define ∞ −∞ f(x)dx=L 1+L 2.

We show that there is a connection with the Lebesgue integral with respectλ.

Theorem 6.12.Le tf:R→Rbe a continuous function. Then the fol lowing are equivalent:

(i)fis integrable; (ii) ∞ −∞ |f(x)|dxexists as an improper Riemann integral.

Moreover, in this case ∞ −∞ fdxexists as an improper Riemann integral and (6.5) Rfdλ= ∞ −∞ fdx.

It may happen that ∞ −∞ fdxexists as an improper Riemann integral withoutfbeing integrable (see Exercise6.10). 45The functionfis usually called the density ofν. 28 MEASURE AND INTEGRATION Proof.(i)⇒(ii): First we make a general observation. Letg:R→Rbe a continuous and integrable function. In Example5.16we have seen that t 0g(x)dx= R1[0,t] gdλ. SettingL 1= R1[0,∞] gdλ, we find L 1− t 0 g(x)dx = L 1− R1[0,t] gdλ = R1[t,∞) gdλ ≤ R1[t,∞) |g|dλ.

Now the DCT yields that R1[t,∞) |g|dλ→0ast→∞, and we may conclude that ∞ 0g(x)dx= L 1exists. The part on (−∞,0] goes similarly and we find (6.5)withfreplaced byg.

Now (i)⇒(ii) follows by lettingg=|f|. Moreover, (6.5)forffollows by takingg=f.

(ii)⇒(i): By Example5.16, monotonicity of improper Riemann integrals, R1[−n,n] |f|dλ= n −n |f(x)|dx≤ ∞ −∞ |f(x)|dx=:M<∞.

Therefore, by the MCT, R|f|dλ= lim n→∞ R1[−n,n] |f|dλ≤M, and hencefis integrable. Example6.13.Letf n:[0,∞)→Rbe defined byf n(x)= 1+ xn ne−2x . Below we show that lim n→∞ n 0fn(x)dx= 1. Recall the standard limit 0≤ 1+ xn n↑e xfor everyx∈[0,∞)and thus 0≤1 [0,n] fn↑f,wheref(x)=e −x . Therefore, by Example5.16 n 0 fn(x)dx= [0,∞) 1[0,n] fndλ (MCT)−→ [0,∞) fdλ (6.5)= ∞ 0 f(x)dx= lim t→∞ (1−e −t)=1.

We end this section with a standard application of the DCT to calculus.

Theorem 6.14(Differentiating under the integral sign).Supposef:R×S→Ris such that the fol lowing hold:

(i)fis continuous and differentiable with respect to its first coordinate and for eachy 0∈(a, b) there exists aδ>0and integrable functiong:S→[0,∞]such that | ∂f∂y (y, s)|≤g(s),y∈(y 0−δ, y 0+δ),s∈S.

(ii)s →f(y, s)is integrable with respect toμ.

Then for al ly∈R, d dy Sf(y, s)dμ(s)= S ∂f ∂y(y, s)dμ(s).

Proof.Fixy 0∈Rand letδandgbe as in (i). Leth n∈(0,δ)forn≥1besuchthath n→0. Let φ n:(y 0−δ, y 0+δ)×S→RandF:(y 0−δ, y 0+δ)→Rbe given by φ n(y, s)=f(y+h n,s)−f(y, s) hn andF(y)= Sf(y, s)dμ(s).

Thenφ n(y, s)→ ∂f∂y (y, s)foreachy∈Rands∈S. Therefore, Theorem4.12yields that s →φ n(y, s) is measurable. From the mean value theorem 46 we obtainφ n(y0,s)= ∂f∂y (yn,s)for somey n∈(y 0,y0+δ), and hence|φ n(y0,s)|≤g(s) for alls∈S. It follows that F (y0) = lim n→∞ F(y 0+h n)−F(y 0) hn = lim n→∞ Sφn(y0,s)dμ(s) (DCT)= S ∂f ∂y(y 0,s)dμ(s).

Exercises Exercise6.1.Use convergence theorems to find lim n→∞ Rfndλin each of the following cases:

(a)f n(x)=1 [4,32] (x) 1+ log(4x)n n(use MCT; answer is 2016). 46See [11, Theorem 6.2.3]: ifg:[a, b]→Ris continuous on [a, b] and differentiable on (a, b), then there exists a pointc∈(a, b) such thatg (c)= g(b)−g(a)b−a . MEASURE AND INTEGRATION 29 (b)f n(x)=1 (1,∞) (x)sin(nx) nx 2 (use DCT; answer is 0).

(c)f n(x)=1 [0,1] (x)nx nsin(nx)−n√x+2n 2 (use DCT; answer is− 12 √2).

Exercise6.2.Assumef, f 1,f2,...:S→ Rare measurable functions such that (i) There is a constantM≥0 such that for alln∈N, S|fn|dμ≤M (ii)f n→fpointwise.

Use Fatou’s lemma to show that S|f|dμ≤M.

Exercise6.3.Find lim n→∞∞ m=1 1 m2+n 2.

Hint:Use the counting measure and the DCT.

Exercise6.4.Deduce Corollary6.8from the MCT applied tos n= n j=1 fjforn≥1.

Exercise ∗6.5.Give the details of Remark6.6(use Exercise5.5).

Exercise6.6.Let (f n)n≥1 be a sequence of measurable functions with values inRorCand f n→fpointwise. Assume there exists an integrable functiong:S→[0,∞) such that|f n|≤g.

(a) Show that S|fn−f|dμ→0asn→∞.

Hint:Apply the DCT in a suitable way.

(b) In the real case we know Sfndμ→ Sfdμ. Derive this in the complex case as well.

Hint:Use (a) and Exercise5.11(c).

Exercise ∗6.7.Letf:S→Rbe a measurable function. Defineν:B(R)→[0,∞]byν(B)= μ(f −1 (B)). Prove the following:

(a)νis a measure.

(b) For every measurableg:R→[0,∞] one has Rg(t)dν(t)= Sg(f(s)) dμ(s).

Hint:Argue as in Theorem6.9step 2.

(c) A functiong:R→ Ris integrable with respect toνif and only ifg◦fis integrable with respect toμ. Moreover, in this case Rg(t)dν(t)= Sg(f(s)) dμ(s).

Hint:Argue as in Theorem6.9step 3.

Exercise ∗6.8.Assumef:R→Ris integrable with respect toλ.

(a) Show that for eachy∈Rthe functionx →sin(xy)f(x) is integrable with respect toλ.

Defineg:R→Rby g(y)= Rsin(xy)f(x)dλ(x) (b) Show thatgis continuous.

Hint:Use the sequential characterization of continuity and one of the convergence theorems.

Exercise ∗∗ 6.9.Use induction and Theorem6.14to show that for each integern≥0, ∞ 0 xne−yx dx=n! yn+1 ,y>0.

In particular, settingy= 1 one obtains: ∞ 0xne−x dx=n!.

Exercise ∗∗ 6.10.Letf:[0,∞)→Rbe given byf(x)= sin(x)x ifx = 0 andf(0) = 1.

(a) Show thatfis not integrable with respect toλ.

Hint:Use the estimate|sin(x)|≥ 12on [πn+ 13π, πn+ 23π] for all integersn≥0.

(b) Show that ∞ 0f(x)dxexists as an improper Riemann integral. 47 47Using some smart tricks one could actually show that the integral equals π2. 30 MEASURE AND INTEGRATION 7.L p-spaces In this section (S,A,μ) is measure space. In this section we want to allow the scalar field to be complex as well and we use the notationKfor this. SoK=RorK=C.

Definition 7.1.Fo rp∈[1,∞)let L p(S)= f:S→K:fis measurable and S|f| pdμ <∞ .

Fo rf∈L p(S)let f p:= S|f| pdμ 1p.

Note thatL 1(S) coincides with the set of integrable functionsf:S→K.

If f−g p= 0, then from Proposition5.8(vi) we can concludef=ga.e. However, we would like to havef=g. Therefore, we will identifyfandgwheneverf=ga.e. 48 So one has to be rather careful if one talks aboutf(s) for a certain fixeds∈S. In integration theory this usually does not lead to any problems sincef=ga.e. implies that Efdμ= Egdμfor anyE∈A.

Example7.2.LetS=Rwith the Lebesgue measureλ.Then1 {0} =1 Q= 0 and1 [0,1]\Q =1 [0,1] inL p(R).

7.1.Minkowski and H¨older’s inequalities.Note that for scalarsα∈K, αf p=|α| f p.

Therefore, the next result can be used to show thatL p(S) is a normed vector space.

Proposition 7.3(Minkowski’s inequality 49).Fo r a l lf, g∈L p(S)we havef+g∈L p(S)and f+g p≤ f p+ g p.

Proof.Observe that for alla, b∈[0,∞) one has (See Exercise7.1) (7.1) (a+b) p=inf θ∈(0,1) θ1−p ap+(1−θ) 1−p bp.

It follows from (7.1) that for allθ∈(0,1) and alls∈S, |f(s)+g(s)| p≤(|f(s)|+|g(s)|) p≤θ 1−p |f(s)| p+(1−θ) 1−p |g(s)| p.

Therefore, by monotonicity and linearity of the integral, we obtain that for allθ∈(0,1), S|f+g| pdμ≤θ 1−p S|f| pdμ+(1−θ) 1−p S|g| pdμ.

Stated differently, this says that for allθ∈(0,1), f+g p p≤θ 1−p f p p+(1−θ) 1−p g p p.

Now the result follows by taking the infimum over allθ∈(0,1) and applying (7.1). Another famous inequality which can be proved with the same method is the following.

Proposition 7.4(H¨older’s inequality 50).Le tp, q∈(1,∞)satisfy 511 p+ 1q=1.Iff∈L p(S)and g∈L q(S),thenfg∈L 1(S)and fg 1≤ f p g q.

Proof.See Exercise7.2. 48More precisely, one can build an equivalent relationf∼giff=galmost everywhere and then consider a quotient space. We will use the above imprecise but more intuitive definition.

49Hermann Minkowski (1864-1909) was a German mathematician who worked in geometry. He also was Albert Einstein’s teacher and provided the 4-dimensional mathematical framework for part of Einstein’s relativity theory.

50Otto H¨older (1859-1937) is most famous for this result and for the notion of H¨older continuity of a function.51These exponents are called conjugate exponents. Ifp=2,thenq= 2 and in this case the inequality is known as the Cauchy-Schwarz inequality. MEASURE AND INTEGRATION 31 7.2.Completeness ofL p.

Theorem 7.5(Riesz–Fischer 52).Le tp∈[1,∞).Then(L p(S), · p)is a Banach space. 53 Proof.Let (f k)∞ k=1 be a Cauchy sequence with respect to the norm · pofL p(S). By a standard argument it suffices to show that (f k)∞ k=1 has a subsequence which is convergent inL p(S).

Recursively, we can find a subsequence (f kn)n≥1 such that (7.2) f kn+1 −f kn p≤1 2n,n=1,2,...

For notational convenience letφ n=f knforn∈N.Alsoletφ 0= 0. We will show that there exists anf∈L p(S) such that φ n−f p→0.

Defineg, g 1,g2,...:S→[0,∞]by g:= ∞ n=0 |φn+1 −φ n|andg m := m−1 n=0 |φn+1 −φ n|,, By Minkowski’s inequality we obtain for eachm≥1, g m p≤ m−1 n=0 φ n+1 −φ n p≤ ∞ n=0 φ n+1 −φ n p(7.2) ≤ φ 1 p+ n≥1 2−n ≤ φ 1 p+1.

Since 0≤g m ↑g, the MCT yields, S|g| pdμ= lim m→∞ S|gm|pdμ= lim m→∞ gm p p≤( φ 1 p+1) p.

LettingA={g<∞}, Exercise5.1yieldsμ(A c) = 0. Therefore, we can definef:S→Rby f= ∞ n=0 1A(φn+1 −φ n), where the series is absolutely convergent andfis measurable by Theorem4.9. By a telescoping argument it follows that pointwise onS f= lim m→∞m−1 n=0 1A(φn+1 −φ n) = lim m→∞ 1Aφm.

Clearly, |φ m|= m−1 n=0 (φn+1 −φ n) ≤ m−1 n=0 |φn+1 −φ n|≤|g| and by lettingm→∞we see that also|f|≤|g|and in particularf∈L p(S). It follows that |f−φ m|p≤(|f|+|φ m|)p≤(2|g|) p=2 p|g| p.

Since|f−φ m|p→0a.e.and2 p|g| pis integrable, it follows from the DCT (see the a.e. version of Remark6.6)that lim m→∞ f−φ m p p= lim m→∞ S|f−φ m|pdμ=0.

In the sequel we will say thatf n→finL p(S)if f n−f p→0.

From the proof of Theorem7.5we deduce the following result.

Corollary 7.6.Le tp∈[1,∞).Supposef, f 1,f2,...∈L p(S).Iff k→finL p(S), then there exists a subsequence(f kn)n≥1 such thatf kn→fa.e. 52Frigyes Riesz (1880–1956) was a Hungarian mathematician who worked in functional analysis. Ernst Fischer (1875–1954) was a Austrian mathematician who worked in analysis.

53Recall that a Banach space is a complete normed vector space. Stefan Banach (1892–1945) was a Polish mathematician who is one of the world’s most important 20th-century mathematicians. He is most famous for his book on functional analysis [2]. 32 MEASURE AND INTEGRATION In general one does not havef n→fa.e. (see Exercise7.7).

Proof.Indeed, since (f n)n≥1 is convergent it is a Cauchy sequence inL p(S). In the proof of Theorem7.5there exists a subsequence (f kn)n≥1 and a function˜ fsuch thatf kn→˜ finL p(S) andf kn→˜ fa.e. Now Minkowski’s inequality yields:

f−˜ f p≤ f−f kn p+ f kn−˜ f p→0.

Thusf=˜ fa.e. and thusf kn→fa.e. Fo rf, g∈L 2(S) we define f, g = Sf(s) g(s)dμ(s), where g(s) stands for the complex conjugate of the numberg(s)∈K.Then ·,· is an inner product and f, f = f 2 2.54 The next theorem is immediate from Theorem7.5.

Theorem 7.7(Riesz–Fischer).L 2(S)is a Hilbert space. 55 Example7.8.Letp∈[1,∞). Ifμ=τis the counting measure onN,then 56 p:=L p(N)coincides with a space of sequences and (a n)n≥1 p= n≥1 |an|p 1p.

Ifp≤q<∞,then p⊆ q(see Exercise7.8).

Example7.9.Assumeμ(S)<∞.If1≤p≤q<∞,thenL q(S)⊆L p(S) (see Exercise7.3).

We end this section with a simple density result. Clearly, the definition of a simple function extends to the complex setting. The following result shows that anyL p-function can be approxi- mated by simple functions.

Proposition 7.10(Density of simple functions).Le tp∈[1,∞). The set of simple functions is dense inL p(S).

Proof.Wr i t ef=u+iv,whereuandvare both real-valued. Writeu=g−h,whereg=u + andh=u −.ByTheorem4.12we can find simple functionsg nandh nsuch that 0≤g n↑g and 0≤h n ↑h.Thenu n :=g n−h nis a simple function,u n →upointwise, and|u n|≤ g n+h n≤g+h≤|u|. Similarly, one constructs simple functionsv nsuch thatv n→vpointwise, |v n|≤|v|. We can conclude thatf n:=u n+iv nis a simple function,f n→fpointwise and |f n|≤(|u n|2+|v n|2)12≤(|u| 2+|v| 2)12≤|f|.Nowsince|f n−f| p≤(|f n|+|f|) p≤2 p|f| pand the latter is integrable, it follows from the DCT that f n−f p→0asn→∞. 7.3.L p-spaces on intervals.In the remaining part of this section we discussL p(I), whereIis an interval andμ=λis the Lebesgue measure onI. These results will not be used in Section8 or in the exercises.

Fo r a n i n t e r va lI⊆Rlet the set of step functions Step(I) be defined by Step(I)=span{1 J:J⊆Iis an interval with finite length}.

One can check that eachφ∈Step(I) is a simple function. The converse does not hold. For instance1 Q∩(0,1) is not a step function. Step function have a lot of structure and often questions onL pcan be reduced to this special class by the following density result. 54Note that the conjugation is need otherwise the square of a complex number can be negative. Physicists often put the conjugation onf(s) instead ofg(s). IfK=R, then the conjugation does not play a role.

55Recall that a Hilbert space is a Banach space where x = x, x 12. David Hilbert (1862–1943) is sometimes said to be the last universal mathematician (which means he knew “all” mathematics of his time). He was one of the most influential mathematicians of the 19th and early 20th centuries. 56Sometimes we wish to use the counting measure onZinstead ofN. In this case we write p(Z):=L p(Z). MEASURE AND INTEGRATION 33 12345 1 2 Figure 7.1.Example step function 141(−1 2,1] + 321(1,5 2]− 131(52,4) +1 [4,5] Theorem 7.11.Le tλbe the Lebesgue measure onI=(a, b)with−∞ ≤a

Proof.Letf∈L p(I)andletε>0. We will construct a step functionφsuch that f−φ p<3ε.

Step 1:Reduction to boundedI.Letf n=1 (−n,n) fforn∈N.Thenf n→fpointwise.

Moreover,|f n−f| p≤|f| p. Therefore, the DCT yields f n−f p→0. Therefore, forn∈Nlarge enough f−g p<ε,whereg=f n∈L p(I).

Step 2:By Step 1 we can assumeIis bounded, and moreover we can assumeI=(a, b]with −∞

Step 3:By Exercise7.5there existF j∈F (a,b] such thatλ(A jF j)< ε n(|x j|+1) p.Nowlet φ= n j=1 1Fjxj. Observe that|1 Aj−1 Fj|=1 Aj F j. Therefore, Minkowski’s inequality yields that h−φ p≤ n j=1 |xj| 1 Aj F j p= n j=1 |xj|λ(A jF j)1/p ≤ n j=1 ε|x j| n(|x j|+1)<ε.

Conclusion:Clearly,φis a step function. Moreover, by Minkowski’s inequality we find f−φ p≤ f−g p+ g−h p+ h−φ p≤3ε.

Corollary 7.12.Le tλbe the Lebesgue measure onI=[a, b]with−∞

Proof.Letf∈L p(I)andletε>0. By Theorem7.11there exists a step functionφsuch that f−φ p<ε. It remains findψ∈C([a, b]) such that φ−ψ p<ε. For this it suffices to approximate an arbitrary1 J. Using Figure7.2and the DCT, the reader can easily convince him or herself that this can indeed be done. 1234 1 2 1234 1 2 1234 1 2 Figure 7.2.Approximation of1 (1, 52]inL pby continuous functions Exercises 34 MEASURE AND INTEGRATION Exercise7.1.Prove the identity (7.1).

Hint:First consider the case wherea, b >0 and minimize the right-hand side of (7.1).

Exercise ∗7.2.Letp, q∈(1,∞) be such that 1p+ 1q=1.

(a) Show that for alla, b≥0, ab=inf t>0 t pap p+b q qtq .

(b) Use the above identity to prove Proposition7.4.

Hint:Argue as in Proposition7.3, but use the identity from (a) instead.

Exercise7.3.Assumeμ(S)<∞and 1≤q≤p<∞. Show thatL p(S)⊆L q(S) and for all f∈L p(S), f q≤μ(S) 1q−1p f p.

Hint:Apply H¨older’s inequality to|f| q·1.

Exercise7.4.Letp∈[1,∞)andλbe the Lebesgue measure onR. Determine for whichα∈R one hasf∈L p(R) in each of the following cases:

(a)f(x)=1 (0,1) (x)x α.

(b)f(x)=1 (1,∞) (x)x α.

Explain whyL p(R) L q(R) for allp, q∈[1,∞)withp =q.

For setsA, B⊆R,AB=(A\B)∪(B\A) denotes thesymmetric differenceofAandB.

Exercise ∗7.5.Letλbe the Lebesgue measure restricted to ((a, b],B((a, b])). LetF (a,b] be the finite unions of half-open intervals in (a, b]. 57 Let A={A∈B((a, b]) :∀ε>0∃F∈F (a,b] such thatλ(AF)<ε}.

(a) ForA, B∈B((a, b]) show thatAB=A cB c.

(b) LetIbe an index set andA i,B i⊆B((a, b]) for alli∈I.LetA= i∈I AiandB= i∈I Bi.

Show thatAB⊆ i∈I AiB i.

(c) Deduce from (a) thatA∈AimpliesA c∈A.

(d) Assume (A n)n≥1 is a disjoint sequence inA. Deduce from (b) that ∞ n=1 An∈A.

Hint:Use that lim n→∞ λ( ∞ k=n Ak) = 0 in order to reduce to finitely many sets.

(e) Show thatA=B((a, b]).

Hint:Use Exercises3.1and3.8.

Exercise ∗7.6.Let−∞

We will derive thatf=0inL 1((a, b]). By considering real and imaginary part separately, one can reduce to the case wherefis real valued.

(a) Show that for allA∈F 1withA⊆(a, b], Afdλ=0.

(b) LetA∈B(R) be such thatA⊆(a, b]. Construct setsA 1,A 2,...∈F 1such thatA k⊆(a, b] and1 An→1 Aa.e.

Hint:Use 1 A−1 B 1=λ(AB) for allA, B∈B(R) and Exercise7.5. Now apply Corollary 7.6.

(c) Show that for allA∈B(R)withA⊆(a, b], Afdλ=0.

Hint:Use (a)and(b) and DCT.

(d) Derive thatf=0 inL 1((a, b]).

Hint:ConsiderA={x∈(a, b]:f(x)≥0}andA={x∈(a, b]:f(x)≤0}. 57By Exercise 1.7 of the lecture notes we can take them disjoint MEASURE AND INTEGRATION 35 Exercise ∗∗ 7.7.Letλbe the Lebesgue measure onR. Observe that eachn∈Ncan be uniquely written asn=2 k+jwithk∈N 0andj∈{0,...,2 k−1}. Now for suchndefinef n = 1 (j2 −k,(j+1)2 −k]. Show thatf n→0inL 1(R), but for allx∈(0,1], there exists infinitely many n∈Nsuch thatf n(x)=1.

Hint:First make a picture fork= 1 andj=0,1, andk= 2 andj=0,1,2,3.

Exercise ∗∗ 7.8.Let 1≤p≤q<∞).

(a) Prove p⊆ qand that for all (a n)n≥1 ∈ pone has (a n)n≥1 q≤ (a n)n≥1 p.

Hint:By homogeneity one can assume (a n)n≥1 p= 1, and therefore|a n|≤1 for alln∈N.

(b) Leta n=n αforn∈N. For whichα∈Rdoesonehave(a n)n≥1 ∈ p?

There is a natural limiting space ofL p(S)forp→∞:

Exercise ∗∗ 7.9.A measurable functionf:S→Kis said to be inL ∞(S) if there exists anM≥0 such thatμ({|f|>M})=0. 58 Define f ∞ =inf{M≥0:μ({|f|>M})=0}.

As usual we identify functionsfandginL ∞(S)iff=ga.e.

(a) Show that (L ∞(S), · ∞) is a Banach space.

(b) Show that for allf∈L∞(S)andg∈L 1(S),fg∈L 1(S)and fg 1≤ f ∞ g 1.

(c) Assumeμ(S)<∞andp∈[1,∞). Show thatL ∞(S)⊆L p(S)and f p≤μ(S) 1p f ∞.

(d) AssumeS=Nwith the counting measure andp∈[1,∞). Let ∞ :=L ∞(N). Show that p⊆ ∞ and (a n)n≥1 ∞ ≤ (a n)n≥1 p.

(e) AssumeIis a finite interval andμ=λis the Lebesgue measure. Show that the simple functions are dense inL ∞(I), but the step functions are not. 58In order words|f|≤Ma.e. 36 MEASURE AND INTEGRATION 8.Applications to Fourier series Fo u r i e r 59 analysis plays a role in a large part of mathematics. In particular, it is one of the central mathematical tools in Physics and Electrical Engineering. A Fourier series is of the form 60 n∈Z cneinx ,or equivalentlya 0 2+ n≥1 ancos(nx)+b nsin(nx), wherex∈[0,2π]. Of course one can also usex∈Rhere. Clearly, the above functions will be periodic whenever they are well-defined.

One of the reasons that Fourier series naturally arise in mathematics is that each of the functions e ±inx ,cos(nx),sin(nx) as an eigenfunctions of d2 dx2with eigenvalue−n 2. Indeed, for instance cos(nx) =−n 2cos(nx).

We have seen that Taylor series can be used to represent functions which are smooth enough. 61 Fourier series provides another tool to represent functions. The class of functions which can be represented as a Fourier series will turn out to be enormous.

In this section we will prove a couple of central results in the theory of Fourier series. The interested reader can read more on the sub ject in [9], [10], [13], [14]and[19]. In particular, very interesting but mostly elementary applications to geometry, ergodicity, number theory and PDEs can be found in [13].

8.1.Fourier coefficients.In this sectionS=[0,2π]andλis the Lebesgue measure on (0,2π).

For notational convenience we will write b a f(x)dx:= [a,b] fdλ.

forf∈L 1(0,2π)and[a, b]⊆[0,2π].

Definition 8.1.Le te k:[0,2π]→Cbe given bye k(x)=e ikx fork∈Z. 62 Le tf∈L 1(0,2π).

(i) Fork∈Zthek-th order Fourier coefficientis defined by 63 f(k)=1 2π 2π 0 f(x) ek(x)dx.

(ii) forn∈N 0,then-th partial sum of the Fourier seriess n(f):[0,2π]→Cis defined by (8.1)s n(f)= |k|≤n f(k)e k.

(iii) A function of the form |k≤n ckekwithn∈N 0and(c k)|k|≤n inCi s ca l l ed atrigonometric polynomial.

Re m a r k8.2.

(1) The reason to use the notion “trigonometric polynomial” is thate k(x)=(e ix)k. Also note thate k(x)=cos(kx)+isin(kx) by Euler’s formula.

(2) Observe that by (the complex version of ) Proposition5.12for eachk∈Z, (8.2)| f(k)|≤1 2π 2π 0 |f(x)e k(x)|dx=1 2π f 1.

In Exercise8.4it will be shown that one even has lim k→∞ f(k)=0.

(3) Iff, g∈L 1(0,2π), the following linearity property holds: (f+g)(k)= f(k)+ g(k) for all k∈Z. This follows directly from the linearity of the integral. 59Fourier analysis is named after the French mathematician Jean-Baptiste Fourier (1768–1830), and were intro- duced in order to solve differential equations such as the heat equation 60Recall Euler’s formula:e ix=cos(x)+isin(x)forx∈R 61In Complex Function Theory these functions will be characterized as the so-called analytic functions62Note thate 0=1 [0,2π] . 63To calculate Fourier transforms numerically one can use the so-called fast Fourier transform FFT (see [13, Section 7.1.3]) MEASURE AND INTEGRATION 37 Example8.3.Givenf∈L 1(0,2π), the functions n(f) is a trigonometric polynomial for each n∈N. Each of the functions cos(nx)=e inx +e −inx 2,sin(nx)=e inx −e −inx 2i is a trigonometric polynomial as well.

Finally note that ifPis a trigonometric polynomial, then for anyj∈N 0,P jis a trigonometric polynomial as well.

The main questions in this section is whether we can reconstructffrom its Fourier coefficients.

More precisely:

•(representation) Which functionsf:[0,2π]→Ccan we write asf= k∈Z ckekfor certain coefficients (c k)k∈Z ?

•(convergence) In what sense does the above series converge?

•(uniqueness) Does f(k)= g(k) for allk∈Zimplyf=g?

When considering convergence of Fourier series we will always consider the convergence of |k|≤n ckek= n k=−n ckek asn→∞.

8.2.Weierstrass’ approximation result and uniqueness.Before we consider convergence of Fourier series, we first we prove a fundamental result about the approximation by trigonometric polynomials. It will be an essential ingredient in the uniqueness result in Theorem8.5.

Theorem 8.4(Weierstrass’ approximation theorem 64 for periodic functions).The trigonometric polynomials are dense in{f∈C([0,2π]) :f(0) =f(2π)}.

Proof.Letf∈C([0,2π]) be such thatf(0) =f(2π)andletε>0 be arbitrary. It suffices to show that there exists a trigonometric polynomialPsuch that f−P ∞ <ε.Weextendfperiodically to a functionf:R→C.Sincefis also uniformly continuous we can chooseδ∈(0,π) such that |x−y|<δimplies|f(x)−f(y)|<ε/2.

Define 65 Fn= 1n n k=1 |j|≤k−1 ejwhich is a periodic function as well. DefineP n:[0,2π]→C byP n(x)= 12π 2π 0 Fn(x−y)f(y)dy.Thensincee j(x−y)=e 2πix e−2πiy the following identity holds P n(x)=1 2π 2π 0 Fn(x−y)f(y)dy=1 2πn n k=1 |j|≤k−1 ej(x) 2π 0 ej(−y)f(y)dy=1 n n k=1 |j|≤k−1 ej(x) f(j).

This shows thatP nis a trigonometric polynomial.

By Exercise8.3the following identity holds F n(z)=sin 2(nz/2) nsin 2(z/2),z∈(0,2π). (8.3) Therefore,F n(z)≥0, and thus we can write F n 1= 2π 0 Fn(z)dz=1 n n k=1 |j|≤k−1 2π 0 ej(z)dz=1 n n k=1 2π=2π. 64Originally Weierstrass proved that the polynomials are dense inC([a, b]). This can be derived from our version of the theorem as indicated in Exercise8.11 65This is called F´ejer’s kernel 38 MEASURE AND INTEGRATION Fixx∈[0,2π]. It follows that 2π|f(x)−P n(x)|= f(x) 2π 0 Fn(x−y)dy− 2π 0 f(y)F n(x−y)dy = 2π 0 Fn(x−y)(f(x)−f(y)) dy ≤ 2π 0 |f(x)−f(y)|F n(x−y)dy ≤T 1+T 2, whereT 1andT 2are the integrals overI:= [x−δ, x+δ]andJ:= [0,x−δ]∪[x+δ,2π], respectively.

On the intervalIwe can use the uniform continuity offto estimate T 1≤ε 2 IFn(x−y)dy≤πε.

OnJ:= [0,x−δ]∪[x+δ,2π] we can estimate T 2≤2 f ∞ JFn(x−y)dy=2 f ∞ BδFn(z)dz, where we substitutedz:=x−yand whereB δ=[δ,2π−δ]. Therefore, using (8.3) again and the fact that|sin(z/2)|≥sin(δ/2) forz∈B δ(recall thatδ≤π) we obtain BδFn(z)dz≤|B δ| nsin 2(δ/2)≤2π nsin 2(δ/2).

So choosingn≥1 so large that 2 f ∞ nsin 2(δ/2) < ε2we obtainT 2<πε.

Therefore, combining the the estimates can conclude that|f(x)−P n(x)|≤ T1+T 2 2π <ε.Since x∈[0,2π] was arbitrary it follows that f−P n ∞ <εas required. Now we can deal with the uniqueness question for Fourier series. This is the most technical part of this section and could be skipped it at first reading.

Theorem 8.5(Uniqueness).Iff∈L 1(0,2π)satisfies f(n)=0for al ln∈Z,thenf=0in L 1(0,2π).

Proof. 66 Step 1:First assumef∈C([0,2π]) andf(0) =f(2π). By linearity it follows that for each trigonometric polynomialPwe have 2π 0 f(x) P(x)dx=0.

By Theorem8.4we can find trigonometric polynomials (P n) such thatP n→funiformly. There- fore, it follows that 2π 0 |f(x)| 2dx= lim n→∞ 2π 0 f(x) Pn(x)dx=0.

This impliesf=0.

Step 2:Next letf∈L 1(0,2π) and assume f(n) = 0 for alln∈Z.LetF:[0,2π]→Rbe defined by F(t)= t 0 f(x)dx.

ThenF∈C([0,2π]) (see Example6.7), andF(0) =F(2π)=0. 67 By Exercise8.6(b) F(k)= f(k)ik =0forallk =0. Nowletg=F−C,whereC= F(0). Theng∈C([0,2π]),g(0) =g(2π) and g(k) = 0 for allk∈Z. Therefore,g= 0 by step 1 and henceF=C.SinceF(0) = 0, this yieldsF(x) = 0 for allx∈[0,2π]. By Exercise7.6we findf=0.

66There is a much better proof in the literature using the F´ejer kernel as an approximate identity. Since approximate identities are not part of these lecture notes, we proceed differently.

67Note thatF(2π)=(2π) f(0) = 0. MEASURE AND INTEGRATION 39 8.3.Fourier series inL 2(0,2π).In this section we will consider Fourier series in the Hilbert spaceL 2(0,2π). Here the inner product is given by f, g = 2π 0 f(x) g(x)dx.

Note thatL 2(0,2π)⊆L 1(0,2π) (see Exercise7.3). Therefore, iff∈L 2(0,2π) the Fourier coeffi- cients are well-defined and f(k)=(2π) −1 f, e k .

Let us recall as special case of Proposition7.3forf, g∈L 2(0,2π), (8.4)| f, g |≤ f 2 g 2 (Cauchy–Schwarz inequality).

We will say thatf, g∈L 2(0,2π) are orthogonal if f, g = 0. Note that in this case the following form of Pythagoras theorem holds 68 (8.5) f+g 2 2= f 2 2+ g 2 2.

Lemma 8.6(Orthogonality).Fo rj, k∈Z, e j,ek = 2π,ifj=k; 0,ifj =k.

Consequently, if finitely many(c j)j∈Z inCare nonzero, then (8.6) j∈Z cjej 2=(2π) 12 j∈Z |cj|2 12.

Proof.Indeed, ifj =k, then using ek(x)=e −ikx we find that e j,ek = 2π 0 ei(j−k)x dx= 2π 0 cos((j−k)x)dx+i 2π 0 sin((j−k)x)dx=0 by periodicity of cos and sin. Similarly, one sees e j,ej =2π.

The final statement follows from j∈Z cjej 2 2= j∈Z k∈Z cjck ej,ek =2π j∈Z |cj|2.

We extend this result to series using the completeness ofL 2(0,2π).

Theorem 8.7(Riesz–Fischer, Convergence of Fourier series inL 2).

(i) If(c n)n∈Z ∈ 2,theng:= n∈Z cnenconverges inL 2(0,2π),and g(n)=c nfor al ln∈Z,and (8.7) g 2=(2π) 12 (c n)n∈Z 2 (Parseval’s identity) (ii) Iff∈L 2(0,2π),then( f(n)) n∈Z in 2andf= n∈Z f(n)e ninL 2(0,2π)and(8.7)holds with g=fandc n= f(n)forn∈Z.

Part (ii) shows that everyL 2-function can be represented as a Fourier series. A similar result holds for series of sine and cosine functions and can be derived as a consequence of the above result (see Exercise8.8).

Proof.(i): Letg n= |k|≤n ckekforn∈N. We show that (g n)n≥1 is a Cauchy sequence. Let ε>0andchooseN∈Nsuch that |k|≥N |ck|2 12<ε (2π) 12. 68This follows by writing out f+g 2 2= f+g, f+g . 40 MEASURE AND INTEGRATION Then for all integersn>m≥Nby Lemma8.6, g n−g m 2= m<|k|≤n ckek 2=(2π) 12 m<|k|≤n |ck|2 12≤(2π) 12 |k|≥N |ck|2 12<ε.

This proves that (g n)n≥1 is a Cauchy sequence. By completeness (see Theorem7.7),g:= lim n→∞ gnexists inL 2(0,2π).

To c h e c k (8.7) note that by the continuity of · 2and Lemma8.6, g 2= lim n→∞ gn 2= lim n→∞ (2π) 12 |k|≤n |ck|2 12=(2π) 12 (c n) 2.

Finally, note that g(k)=(2π) −1 g, e k = lim n→∞ (2π) −1 gn,ek =c k.69 (ii): Fixn∈N.Since f−s n(f),e k = 0 for each|k|≤n,also f−s n(f),s n(f) = 0 and hence (8.5) yields (8.8) f 2 2= f−s n(f)+s n(f) 2 2= f−s n(f) 2 2+ s n(f) 2 2≥ s n(f) 2 2.

Letc k= f(k)fork∈Z.Then(8.8) yields:

f 2 2≥ s n(f) 2(8.6) = |k|≤n 2π|c k|2.

Lettingn→∞, we find (c k)k∈Z 2≤(2π) −1 f 2<∞.

By (i) we can defineg= n∈Z cnenwhere the series converges inL 2(0,2π). We claim that f=ginL 2(0,2π). To see this note that by (i), g(n)=c n= f(n). Therefore, the claim follows from the uniqueness Theorem8.5applied tof−g. Fo rf∈L 2(0,2π)Theorem8.7yields thatf−s n(f)= |k|>n f(k)e kand by (8.7) (8.9) (L 2-error estimate) f−s n(f) 2 2=2π |k|>n | f(k)| 2.

Moreover, sinces n(f) is a trigonometric polynomial, we also find the following: 70 Corollary 8.8.The trigonometric polynomials are dense inL 2(0,2π).

Example8.9 (Sawtooth function).Letf:[0,2π)→Rbe defined byf(x)=x−πand extended periodically onR.Fork∈Z\{0}, by the Fundamental Theorem of Calculus (see [11,Theorem 7.3.1]) and integration by parts, (8.10) f(k)=1 2π 2π 0 (x−π)e −ikx dx=1 2π (x−π)e −ikx −ik 2π 0 −1 2π 2π 0 e−ikx −ikdx=−1 ik.

Clearly, f(0) = 0. Therefore, Theorem8.7yield thatf=− k∈Z\{0}e kik with convergence in L 2(0,2π). Moreover, using that 2 sin(kx)= ek−e−ki we also find thatf=−2 ∞ k=1sin(k·) k with convergence inL 2(0,2π) (see Figure8.1) for plots of the partial sums).

TheL 2-error can be estimated using (8.9):

f−s n(f) 2 2=2π |k|>n | f(k)| 2≤4π k≥n+1 1 k2≤4π ∞ n 1 x2dx=4π n.

One can show thats n(f)willnotconvergetofuniformly (also see Figure8.1). This is in particular clear forx= 0 andx=2π, becausef(0) =πandf(2π)=−π, buts n(f)(0) =s n(f)(2π)=0.

By applying (8.7) one can obtain a remarkable identity:

f 2 2=2π k∈Z\{0} 1 k2. 69Here we used| g−g n,ek | ≤ g−g n 2 ek 2as follows from (8.5). 70With more advanced techniques one can show that the trigonometric polynomials are dense in anyL p(0,2π) withp∈[1,∞). MEASURE AND INTEGRATION 41 On the other hand, if we calculate f 2 2with the fundamental theorem of calculus, we obtain f 2 2= 2π 0 (x−π) 2dx= 13(x−π) 3 2π 0 =2 3π 3.

Combining both identities gives k∈Z\{0}1 k2= 13π2, and so we find ∞ k=11 k2= 16π2. −10−5510 −4 −2 2 4 −10−5510 −4 −2 2 4 −10−5510 −4 −2 2 4 Figure 8.1.The Fourier series of the sawtooth function withn=2,n= 5 and n= 10.

8.4.Fourier series inC([0,2π]).In this section we will give some sufficient condition onfwhich imply the Fourier series is uniformly convergent (or equivalently convergent inC([0,2π]) with the supremum norm · ∞). Note thatf n→funiformly implies thatf n→finL 2(0,2π). Indeed, this follows from (8.11) f n−f 2 2≤ 2π 0 |fn(x)−f(x)| 2dx≤ f n−f 2 ∞ 2π 0 1dx=2π f n−f 2 ∞.

From the above we see that convergence of Fouier series inC([0,2π]) is stronger than conver- gence inL 2(0,2π). However, there are example of functionsf∈C([0,2π]) withf(0) =f(2π)for which the uniform convergence (and even the pointwise convergence) fails (see [1, Example 35.11] and [10, Example 2.5.1]). So apparently more restrictive conditions are needed.

All the different types of convergence can be confusing. Let us summarize some convergence results for a sequence (f n)n≥1 inL 2(0,2π). 71 uniform conv.L 2-conv. =⇒L 1-conv. =⇒a.e.-conv. subsequence. =⇒ =⇒ pointwise conv. =⇒a.e.-conv.

L 1andL 2-convergence are only implied by a.e. conv. under additional assumptions on the (f n)n≥1 (as given for instance in the DCT).

The following result provides sufficient conditions for uniform convergence.

Theorem 8.10(Absolute and uniform convergence of Fourier series).Le tf∈C([0,2π]).If ( f(k)) k∈Z ∈ 1,thenf(x)= k∈Z f(k)e k(x),x∈[0,2π]where the series is absolutely and uniformly convergent.

As a consequence we see thatf(0) =f(2π) holds in this situation, becausee k(0) =e k(2π).

Proof.Fo r a l lx∈[0,2π], k∈Z | f(k)e k(x)|= k∈Z | f(k)|<∞. 71For completeness we note that a.e.-conv. =⇒conv. in measure =⇒conv. in distribution. 42 MEASURE AND INTEGRATION Therefore, we can letg= k∈Z f(k)e k, where the series is absolutely convergent. Moreover, g−s n(f) ∞ ≤ |k|>n | f(k)|→0, and henceg∈C([0,2π]) ands n(f)→guniformly. By (8.11) the convergence holds inL 2(0,2π) as well, and hence 2π g(k)= g, e k = lim n→∞ sn(f),e k =2π f(k) and therefore,g=fa.e. by Theorem8.5.LetA={s∈[0,2π]:f(s)=g(s)}.ThenAis closed andλ(A)=2π. We claim thatAis dense. Indeed, if not then there exists an nonempty open intervalI⊆[0,2π]\A. It follows that 0<λ(I)≤λ([0,2π]\A)=λ([0,2π])−λ(A) = 0. This is a contradiction and thus the claim follows. SinceAis also closed in [0,2π], the claim implies that A=[0,2π]. From the proof we see that the following error estimate holds:

(8.12) (uniform error estimate) f− sn(f) ∞≤ |k|>n | f(k)|.

The condition of Theorem8.10holds in the following situation:

Corollary 8.11.Assumef∈L 2(0,2π)satisfies f(0) = 0.Supposec 0∈CandF:[0,2π]→Kis given byF(t)=c 0+ t 0f(x)dx.Then( F(k)) k∈Z ∈ 1andF= k∈Z F(k)e kwhere the series is absolutely and uniformly convergent.

Proof.Sincef∈L 2(0,2π)⊆L 1(0,2π), it follows from Example6.7thatFis continuous on [0,2π].

Moreover, for everyt∈[0,2π], |F(t)|≤|c 0|+ 2π 0 |f(x)|dx≤|c 0|+ f 1.

In particular, by monotonicity we see that F 2≤(2π) 12(|c 0|+ f 1).

By Exercise8.6, F(k)= f(k)ik fork =0. ByTheorem8.7, f(k)) k∈Z 2= f 2. Therefore, by the Cauchy–Schwarz inequality (8.4) k∈Z | F(k)|=| F(0)|+ k∈Z\{0} | f(k)| |k|≤| F(0)|+C ( f(k)) k∈Z 2≤| F(0)|+C (2π) 12 f 2, where used (8.7) in the last step. Therefore, the absolute and uniform convergence follows from Theorem8.10. Example8.12.Ifg∈C([0,2π]) satisfiesg(0) =g(2π),gis piecewise continuously differentiable on (0,2π)andg ∈L 2(0,2π), theng= k∈Z g(k)e kwhere the series is absolutely and uniformly convergent. Indeed, letF=g−g(0). ThenF(t)=−g(0) + t 0g (x)dx,wheref:=g satisfies the assumptions of Corollary8.11.

Exercises Exercise8.1.Letf:[0,2π)→Rbe given byf=1 [0,π] .

(a) Show that f(k) = 0 for evenk =0, and f(k)= 1πik for oddk,and f(0) = 12.

(b) Writefas a series of sines as in Example8.9and give an estimate ofL 2-error given by (8.9).

(c) Evaluate j∈Z1 (2j+1) 2.

Hint:Argue as in Example8.9.

Exercise8.2.Letf:[0,2π)→Rbe given byf(x)=|x−π|. Example8.12yields that the Fourier series offis absolutely and uniformly convergent. Calculate the Fourier series explicitly and give estimates for theL 2-error (8.9) and uniform error (8.12). One can also obtain the exact form of another famous series k∈Z | f(k)| 2using Parseval and the fundamental theorem of calculus. MEASURE AND INTEGRATION 43 Exercise8.3 (Special Fourier series and kernels).The following kernel’s play a central role in more advanced theory of Fourier series. Prove the identities below forx∈(0,2π). For each exercise you should use the geometric sum n k=0 ak= 1−a n+1 1−a fora∈C\{1}.

(a) (Dirichlet kernel) Show thatD n(x):= |k|≤n−1 ek(x)=sin((n− 12)x) sin( 12x)forn≥1.

(b) (F´ejer kernel) Show thatF n(x):=1 n n j=1 Dj(x)=1 nsin 2(n x2) sin 2(12x)forn≥1.

Exercise ∗8.4 (Riemann-Lebesgue lemma).

(a) Show that for any step functionf:[0,2π]→C(see Section7.3) one has lim |k|→∞ f(k)=0.

Hint:By linearity it suffices to considerf=1 (a,b) ,where(a, b)⊆[0,2π].

(b) Show that for anyf∈L 1(0,2π) one has lim |k|→∞ f(k)=0.

Hint:Use Theorem7.11and (a).

Exercise ∗8.5.

(a) Let ( H, ·,· ) be a Hilbert space (over the complex scalars). Prove that for allu, v∈H, (polarization) 4 u, v = u+v 2− u−v 2+i u+iv 2−i u−iv 2.

(b) Use (a)and(8.7)toprovethatforallf, g∈L 2(0,2π):

2π 0 f(x) g(x)dx=2π k∈Z f(k) g(k).

Exercise ∗8.6.Letg∈C 1([0,2π]) 72 andf∈L 1(0,2π). DefineF:[0,2π]→CbyF(t)= t 0f(x)dx. By Example6.7,Fis continuous.

(a) Prove the following integration by parts formula:

2π 0 f(x)g(x)dx=F(2π)g(2π)−F(0)g(0)− 2π 0 F(x)g (x)dx.

Hint:For continuousfthis is just the standard integration by parts formula. Use Corollary 7.12and approximation to deduce the general case.

(b) Show that f(k)= f(0) +ik F(k) for allk∈Z\{0}.

Hint:Apply (a)withg=e −k .

Exercise ∗8.7.AssumeF:[0,2π]→Cis continuously differentiable and satisfiesF(0) =F(2π) and F(0) = 0. Letf=F .

(a) Show that f(k)=ik F(k) for allk∈Z.

Hint:Apply Exercise8.6(b).

(b) Show that F 2≤ f 2and that equality holds if and only ifF=c 1e1+c −1 e−1 forc 1,c−1 ∈C.

Hint:Apply (8.7).

Exercise ∗8.8.Consider Γ ={ 121[0,2π] }∪{cos(n·):n∈N}∪{sin(n·):n∈N}⊆L 2(0,2π).

(a) Show thatφ, ψ∈Γwithφ =ψare orthogonal and φ 2=π.

(b) Letf∈L 1(0,2π)besuchthat f, φ =0 for allφ∈Γ. Show thatf=0.

Hint:Use Theorem8.5 (c) Show that for every (a n)n≥0 ,(b n)n≥1 ∈ 2the following series converges inL 2(0,2π).

g:=a 0 2+ n≥1 ancos(n·)+ n≥1 bnsin(n·) Hint:Argue as in Theorem8.7.

(d) Show that g 2 L2(0,2π) =π n≥0 |an|2+π n≥1 |bn|2.

Hint:Argue as in Theorem8.7. 72That meansgis differentiable and its derivative is continuous on [0,2π] 44 MEASURE AND INTEGRATION (e) Show that (a n)n≥0 and (b n)n≥1 satisfy a n=1 π 2π 0 cos(nx)g(x)dx, b n=1 π 2π 0 sin(nx)g(x)dx.

Hint:Argue as in Theorem8.7.

(f ) Show that everyf∈L 2(0,2π) can be written as f=a 0 2+ n≥1 ancos(n·)+b nsin(n·) with converges inL 2(0,2π).

Hint:Apply Theorem8.7or argue as in Theorem8.7.

It follows from the previous exercise and (8.2)thatforC 1-functionsFone has| F(k)|≤ f 1 2π|k| for allk∈Z\{0}. Moreover, in Exercise8.4we have seen that for generalF∈L 1(0,2π) one has F(k)→0as|k|→∞. In the next exercise we show that the convergence can be arbitrary slow even for periodic functionsF∈C([0,2π]).

Exercise ∗∗ 8.9.Show that for any sequence (c k)k≥1 withc k = 0 andc k→0 there exists a functionF∈C([0,2π]) withF(0) =F(2π) such that| F(k)|≥|c k|for infinitely manyk∈N.

Hint:Choose a subsequence such that ∞ n=1 |ckn|<∞.

Finally we deduce Weierstrass’ classical approximation result. We first need an elementary result about even trigonometric polynomials.

Exercise ∗∗ 8.10.LetP= n k=−n ckekbe a trigonometric polynomial.

(a) Show that there exists a polynomialq nof degreensuch that cos(nx)=q n(cos(x)).

Hint:Use induction and the recursion formula cos(kx) + cos((k−2)x) = 2 cos((k−1)x) cos(x).

(b) Show that there exists a polynomialq nof degreensuch that sin(nx)sin(x) =r n(cos(x)).

Hint:Use induction and the recursion formula sin((k+1)x)+sin((k−1)x) = 2 cos(kx)sin(x).

(c) From (a)and(b) derive that there there exit polynomialsq, r:[−π, π]→Rof degreensuch thatP(x)=q(cos(x)) +r(cos(x)) sin(x).

(d) If additionallyPis even (i.e.P(−x)=P(x)forx∈[−π, π]). Show that there exists a polynomialqof degreensuch thatP(x)=q(cos(x)).

Hint:Wr i t e 2P(x)=P(x)+P(−x) and use (c).

Exercise ∗∗ 8.11 (Weierstrass’ approximation theorem for continuous functions).Letf:[−1,1]→ Rbe continuous and letε>0. Letg:[−π, π]→Rgiven byg(x)=f(cos(x)).

(a) Show that there is aneventrigonometric polynomialPsuch that g−P ∞ <ε.

Hint:First apply Theorem8.4on [−π, π] to obtain a trigonometric polynomialPsuch that g−P ∞ <ε. Now consider P(x)= P(−x)+P(x) 2 .

(b) Use Exercise8.10(d) to find a a polynomialqsuch that f−q ∞ <ε.

(c) The above shows that the polynomials are dense inC([−1,1]). Use a scaling argument to show that the polynomials are dense inC([a, b]). MEASURE AND INTEGRATION 45 Final remarks:

•One can show that f−F n∗f p→0 for allf∈L p(0,2π)withp∈[1,∞). In Theo- rem8.4The convergence holds uniformly iff∈C([0,2π]) satisfiesf(0) =f(2π). Here F n∗fis the so-called convolution product ofF nandfandisdefinedbyF n∗f(t)= 2π 0 Fn(t−x)f(x)dx. Using the definition ofF none can check thatF n∗fis a trigono- metric polynomial.

•Similar results forD naretrueaswellaslongasp∈(1,∞), but this a much deeper result. SinceD n∗f=s n(f), this implies that f−s n(f) p→0 for allf∈L p(0,2π)with p∈(1,∞).

•Finally, we note thats n(f)→fa.e. for anyf∈L p(0,2π)withp>1. This is one of the deepest result in the theory of Fourier series and was proved by Carleson forp= 2 and extended top>1 by Hunt in 1968. It was proved a long time before that the result fails forp= 1 by Kolmogorov in 1923.

For details we refer to the elective Bachelor course on Fourier analysis ! 46 MEASURE AND INTEGRATION AppendixA.Dynkin’s lemma The results of this section are not part of the exam material. We will proof the uniqueness result of Proposition3.5. In this sectionSdenotes a set.

Definition A.1(π-system).A col lectionE⊆P(S)i s ca l l ed aπ-systemif for al lA, B∈Eone hasA∩B∈E.

ExampleA.2.Every ring is aπ-system.

Definition A.3.A col lectionD⊆P(S)i s ca l l ed aDynkin-system 73 if the fol lowing conditions hold:

(D1)S∈D; (D2)A, B∈DandA⊆BimpliesB\A∈D; (D3) If(A n)n≥1 inDand 74 An↑A,thenA∈D.

ExampleA.4.LetS={1,2,3,4}andD={∅,S,{1,2},{3,4},{1,3},{2,4}}.ThenDis a Dynkin system, but it is not aπ-system.

Proposition A.5.Fo r a co l l ec t i o nF⊆P(S)the fol lowing are equivalent:

(i)Fis aσ-algebra; (ii)Fis a Dynkin system and aπ-system.

Proof.(i)⇒(ii): This is ExerciseA.1.

(ii)⇒(i):∅,S∈Ffollows from (D1) and (D2) of the definition of a Dynkin system. Let (A n)n≥1 be a sequence inF.LetA= ∞ j=1 AjandB n= n j=1 Ajforn≥1. SinceFis aπ-system it is closed under finite intersections. Therefore, using (D2) we obtainB n= n j=1 Ac j c∈F.Since B n↑A,itfollowsfrom(D3)thatA∈F. Lemma A.6(Dynkin).Le tE⊆P(S)be aπ-system andD⊆P(S)be a Dynkin system. If E⊆D,thenσ(E)⊆D.

Proof.LetD 0denote the intersection of all Dynkin system which containE. Observation:E⊆ D 0⊆DandD 0is a Dynkin system (see ExerciseA.2). We claim thatD 0is aπ-system as well.

As soon as we have proved this claim, PropositionA.5yields thatD 0is aσ-algebra. Therefore, from the observation it follows thatσ(E)⊆D 0⊆D. To prove the claim we need two steps.

Step 1:Define a new collection by D 1={D∈D 0:D∩E∈D 0 for eachE∈E}.

SinceEis aπ-system alsoE⊆D 1. The collectionD 1is a Dynkin-system again. Indeed,S∈D 1is clear. IfA, B∈D 1andB⊆A,thenforeachE∈Ewe find (B\A)∩E=(B∩E)\(A∩E)∈D 0, becauseA∩E, B∩E∈D 0andD 0is aπ-system, and thusB\A∈D 1. Next let (A n)n≥1 inD 1 withA n↑A.ThenforeachE∈E,A∩E= ∞ n=1 (A n∩E)∈D 0sinceA n∩E∈D 0.SinceD 0⊆D 1 andD 1is a Dynkin system which containsE, we findD 1=D 0.

Step 2:Define a new collection by D 2={D∈D 0:D∩C∈D 0 for eachC∈D 0}.

SinceD 1=D 0, we find thatE⊆D 2. As before one checks thatD 2is a Dynkin system. Moreover, as before this yieldsD 2=D 0. Thisprovestheclaim. Proposition A.7(Uniqueness).Le tμ 1andμ 2both be measures on measurable space(S,A).

Assume the fol lowing conditions:

(i)E⊆Ais aπ-system withσ(E)=A; 73Eugene Dynkin 1924–2014 was a Russian mathematician who worked on Algebra and Probability theory.74See Definition2.9for the meaning ofA n↑A MEASURE AND INTEGRATION 47 (ii)μ 1(S)=μ 2(S)<∞andμ 1(E)=μ 2(E)for al lE∈E.

Thenμ 1=μ 2onA.

Proof.LetD={A∈A:μ 1(A)=μ 2(A)}.ThenE⊆D. We claim thatDis a Dynkin system.

From the claim and LemmaA.6it follows thatA=σ(E)⊆D⊆A. This impliesD=Aand the required result follows from the Definition ofD.

To prove the claim note thatS∈Dby assumption. IfA, B∈DwithA⊆B,thenμ 1(B\A)= μ 1(B)−μ 1(A)=μ 2(B)−μ 2(A)=μ 2(B\A) and henceB\A∈D. Finally, if (A n)n≥1 inDwith A n↑A, then by Theorem2.10,μ j(A n)↑μ j(A)forj=0,1. Sinceμ 1(A n)=μ 2(A n), this yields μ 1(A)=μ 2(A) and thusA∈D. Exercises ExerciseA.1.Prove PropositionA.5(i)⇒(ii).

Exercise ∗A.2.Prove that the intersection of Dynkin systems is again a Dynkin system.

Exercise ∗A.3.Find a version of PropositionA.7which for measures withμ 1(S)=μ 2(S)=∞.

Hint:See the proof of Theorem3.10. 48 MEASURE AND INTEGRATION AppendixB.Carath´ eodory’s extension theorem The results of this section are not part of the exam material (although the statement of the Carath´eodory’s extension in Theorem3.1is part of the exam).

In order to state and prove Carath´eodory extension Theorem3.1we first a new concept of measurability associated to a mappingα:P→[0,∞] which satisfiesα(∅)=0.

Definition B.1.Le tα:P(S)→[0,∞]be a mapping which satisfiesα(∅)=0. We say that A⊆Sisα-measurable if α(Q)=α(Q∩A)+α(Q∩A c),for al lQ∈P(S).

The col lection of al lα-measurable sets is denoted byM α.

The mappingαcould be rather general and it will not be additive in general. However, the cleverly defined collectionM αturns out to be a ring on whichαis additive.

Lemma B.2.Le tα:P(S)→[0,∞]be a mapping which satisfiesα(∅)=0.ThenM αis a ring andαis additive onM α.

Proof.In order to check thatM αis a ring we check the following:

(i)∅∈M α; (ii)A∈M α=⇒A c∈M α; (iii)A, B∈M α=⇒A∩B∈M α.

Given these properties it is straightforward to check thatM αis a ring. Indeed, this follows from the formulasB\A=B∩A candA∪B=(A c∩B c)c.

Properties (i) and (ii) are clear. It remains to check (iii). LetA, B∈M αand writeC=A∩B.

LetQ∈P(S) be arbitrary. Observation:A∩B c=C c∩AandA c=C c∩A c. One has α(Q)=α(Q∩A)+α(Q∩A c)sinceA∈M α =α(Q∩A∩B)+α(Q∩A∩B c)+α(Q∩A c)sinceB∈M α =α(Q∩C)+α(Q∩C c∩A)+α(Q∩C c∩A c) by the observation =α(Q∩C)+α(Q∩C c)sinceA∈M α.

Therefore,A∩B=C∈M αwhich yields (iii).

To c h e c k t h a tαis additive fix two disjoint setsA, B∈M αand letQ=A∪B.ThenQ∩A=A andQ∩A c=B. Therefore, sinceA∈M αwe find α(A∪B)=α(Q)=α(Q∩A)+α(Q∩A c)=α(A)+α(B).

In LemmaB.2we have seen thatαis additive. In order to obtain a measure we need a further condition onα.

Definition B.3.Le tSbe a set. A functionα:P(S)→[0,∞]is cal led anouter measureif (i)α(∅)=0; (ii) (monotonicity)A⊆B=⇒α(A)≤α(B).

(iii) (σ-subadditivity) For each sequence(A n)n≥1 inP(S)one hasα ∞ n=1 An ≤ ∞ n=1 α(A n).

An outer measure is not necessarily a measure. For instanceα(∅)=0andα(A)=1ifA⊆S is non-empty is an example of an outer measure which is not a measure (ifScontains at least two elements).

Next, we show that being an outer measure is the right additional ingredient to prove thatα is a measure onM α.

Lemma B.4.Le tαbe an outer measure. ThenM αis aσ-algebra andαis a measure on(S,M α). MEASURE AND INTEGRATION 49 Proof.LemmaB.2gives thatM αis a ring andαis additive onM α. It remains to check that for any disjoint sequence (A n)n≥1 , (B.1)A:= ∞ n=1 An∈M α andα(A)= ∞ n=1 α(A n).

LetB n= n j=1 Ajfor eachn≥1. Fix an arbitraryQ⊆S. For everyn∈Nthe following holds:

n j=1 α(Q∩A j)+α(Q∩A c)=α(Q∩B n)+α(Q∩A c) (by LemmaB.2) ≤α(Q∩B n)+α(Q∩B c n)(sinceA c⊆B c n) =α(Q)(sinceB n∈M α) Using first theσ-subadditivity ofαand then the arbitrariness ofn∈Nintheabove,wededuce (B.2)α(Q∩A)+α(Q∩A c)≤ ∞ j=1 α(Q∩A j)+α(Q∩A c)≤α(Q).

On the other hand by subadditivity also the converse estimate holds:α(Q)≤α(Q∩A)+α(Q∩A c), and henceA∈M α. Moreover, all inequalities in (B.2) have to be identities 75, and hence (B.1) follows by takingQ=Ain (B.2). In the above results we have seen that with outer measures one can construct measures on certainσ-algebras. Our next aim is to show that there is a natural outer measure associated to an additive mappingμon a ringR.

Lemma B.5.Le tSbe a set andR⊆P(S)be a ring. Supposeμ:R→[0,∞]is additive and satisfiesμ(∅)=0.ForA⊆Sdefine (B.3)μ ∗(A)=inf ∞ j=1 μ(B j):A⊆ ∞ j=1 Bj,whereB j∈Rforj≥1 , where we letμ ∗(A)=∞if the above set is empty. Thenμ ∗is an outer measure Proof.The mappingμ ∗:P(S)→[0,∞] clearly satisfies (i) and (ii). In order to check (iii) let (A n)n≥1 inPand letε>0. Ifμ ∗(A n)=∞for somen≥1, then (iii) is trivial. Next assume μ ∗(A n)<∞for alln≥1. Then by definition ofμ ∗for each fixedn≥1 we can findB n,j ∈R such that A n⊆ ∞ j=1 Bn,j andμ ∗(A n)+2 −n ε≥ ∞ j=1 μ(B n,j ).

Then ∞ n=1 An⊆ ∞ n,j=1 Bn,j and again by the definition ofμ ∗we find μ ∗ ∞ n=1 An ≤ ∞ n,j=1 μ(B n,j )≤ ∞ n=1 μ∗(A n)+2 −n ε=ε+ ∞ n=1 μ∗(A n).

Sinceε>0 was arbitrary, this prove the required estimate. Theorem B.6(Carath´eodory’s extension theorem).Le tR⊆P(S)be a ring andμ:R→[0,∞] beσ-additive onRandμ(∅)=0.Thenμ ∗defined by(B.3)satisfies the fol lowing properties:

(i)μ ∗is a measure on theσ-algebraM μ∗; (ii)μ ∗(A)=μ(A)for al lA∈R.

(iii)R⊆M μ∗; In particular, μ:σ(R)→[0,∞]defined by μ(A)=μ ∗(A)defines a measure.

Clearly, Theorem3.1follows from the above statement and actually shows that there is a further extension to the possibly largerσ-algebraM μ∗. 75Clearly,x≤y≤z≤xenforcesx=y=z 50 MEASURE AND INTEGRATION Proof.In LemmaB.5we have proved thatμ ∗is an outer measure. Therefore, assertion (i) follows from LemmaB.4. In remains to prove (ii) and (iii). The assertion concerning μfollows from (i)–(iii) as the restriction of a measure to a smallerσ-algebra is a measure again.

Step 1: Proof of (ii):LetA∈R. It is clear thatμ ∗(A)≤μ(A). Indeed, takeB 1=Aand B n=∅forn≥2in(B.3). For the converse estimate the caseμ ∗(A)=∞is clear. Now suppose μ ∗(A)<∞and letB 1,B 2,...,∈Rbe such thatA⊆ ∞ n=1 Bn. Then by theσ-additivity ofμon Rand Theorem2.8(iii) applied toA= ∞ n=1 A∩B n, we find μ(A)≤ ∞ n=1 μ(A∩B n)≤ ∞ n=1 μ(B n).

Taking the infimum over all (B n)n≥1 as above yieldsμ(A)≤μ ∗(A).

Step 2: Proof of (iii):LetA∈RandQ⊆S.Sinceμ ∗is subadditive we findμ ∗(Q)≤ μ ∗(Q∩A)+μ ∗(Q∩A c). For the converse estimate the caseμ ∗(Q)=∞is trivial. In caseμ ∗(Q)< ∞, chooseB 1,B 2,...,∈Rsuch thatQ⊆ ∞ n=1 Bn.ThenB n∩A, B n∩A c∈Rfor alln≥1and Q∩A⊆ ∞ n=1 Bn∩AandQ∩A c⊆ ∞ n=1 Bn∩A c.

Therefore, using first the definition ofμ ∗and then the additivity ofμonR, we find μ ∗(Q∩A)+μ ∗(Q∩A c)≤ ∞ n=1 μ(B n∩A)+ ∞ n=1 μ(B n∩A c)= ∞ n=1 μ(B n) Taking the infimum over all (B n)n≥1 as above givesμ ∗(Q∩A)+μ ∗(Q∩A c)≤μ ∗(Q). Combining both estimates we can concludeA∈M μ∗. Exercises In the following exercise we show thatM μ∗ =P(S) in general. 76 ExerciseB.1.LetS={1,2,3}and define aσ-algebra byA={∅,S,{1,2},{3}}. Assumeμis a measure satisfyingμ({1,2})=μ({3})= 12.

(a) Show thatμ ∗({1})=μ ∗({2})= 12.

(b) Show that{1},{2}/ ∈M μ∗.

ExerciseB.2.Letα:P(S)→[0,∞] be an outer measure and suppose thatA⊆P(S) satisfies α(A) = 0. Show thatA∈M α.

Exercise ∗B.3.Assume the conditions of TheoremB.6and assumeμisσ-finite onR, that means there exists a sequence (S n)n≥1 inRsuch thatμ(S n)<∞for alln≥1and ∞ n=1 Sn=S.Prove that the following are equivalent:

(a)A∈M μ∗; (b) There exists aB∈σ(R) such thatA⊆Bandμ ∗(B\A)=0 Hint:First reduce to the case of finite measure by intersecting withS n. Use the definition ofμ ∗ givenin(B.3). 76For the Lebesgue measure one also hasM μ∗ =P(R), but this is much harder to prove. See AppendixC MEASURE AND INTEGRATION 51 AppendixC.Non-measurable sets Letλbe the Lebesgue measure onB(R d). Letλ ∗:P(S)→[0,∞] be the outer measure associated withλ(see LemmaB.5). Theσ-algebraM(Rd):=M λ∗introduced in DefinitionB.1 is usually called theLebesgueσ-algebra. It follows from TheoremB.6thatF d⊆M(R d)and thus alsoB(R d)⊆M(R d)andλ ∗is a measure onM(R d). In the sequel we writeλagain for this measure as it is just an extension ofλ.

By ExerciseB.3for everyA∈M(R d)thereexistsaB∈B(R d) such thatA⊆Bandλ(B\A)= 0. This shows that the Lebesgueσ-algebraisalmostthesameastheBorelσ-algebra up to sets of measure zero. Some strange things can happen with nonmeasurable sets.

The balls of Banach and Tarski One can cut a ball of radius one inR din such a way that it can be used to form two balls of radius one. Of course something has to be nonmeasurable there. See:

https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox A set which Lebesgue measurable but not Borel measurable There exist a setA∈M(R d)withλ(A) = 0, butA/∈B(R d). See [3, Appendix C] and [12, page 53] http://onlinelibrary.wiley.com/doi/10.1002/9781118032732.app3/pdf http://www.math3ma.com/mathema/2015/8/9/lebesgue-but-not-borel You can also read about this in the Bachelor thesis of Gerrit Vos:

http://resolver.tudelft.nl/uuid:30d69b56-b846-435e-9d44-6a31b840a836 There exist sets which are not Lebesgue measurableA subset ofRwhich is not inM(R d) is given by Vitali’s example (see for example [4, Theorem 16.31]):

https://en.wikipedia.org/wiki/Vitali_set 52 MEASURE AND INTEGRATION References [1] C. D. Aliprantis and O. Burkinshaw.Principles of real analysis. Academic Press, Inc., San Diego, CA, third edition, 1998.

[2] S. Banach.Th´eorie des op´erations lin´eaires.´ Editions Jacques Gabay, Sceaux, 1993. Reprint of the 1932 original.

[3] F. Burk.Lebesgue measure and integration. Pure and Applied Mathematics (New York). John Wiley & Sons, Inc., New York, 1998. An introduction, A Wiley-Interscience Publication.

[4] N. L. Carothers.Real analysis. Cambridge University Press, Cambridge, 2000.

[5] D. H. Fremlin.Measure theory. Vol. 1, 2, 3, 4. Torres Fremlin, Colchester, 2004-2006.

[6] D.J.H. Garling.Inequalities: a journey into linear analysis. Cambridge University Press, Cambridge, 2007.

[7] R. Haberman.Applied Partial Differential Equations with Fourier Series and Boundary Valve Problems.

Pearson Higher Ed, 2012.

[8] O. Kallenberg.Foundations of modern probability. Probability and its Applications (New York). Springer- Verlag, New York, second edition, 2002.

[9] Y. Katznelson.An introduction to harmonic analysis. Cambridge Mathematical Library. Cambridge University Press, Cambridge, third edition, 2004.

[10] J Korevaar. Fourier analysis and related topics.https://staff.fnwi.uva.nl/j.korevaar/Foubook.pdf.

[11] S.R. Lay.Analysis. With an introduction to proof. 5th ed.Pearson, 2015.

[12] H. Royden and P. Fitzpatrick.Real analysis. 4th ed.New York, NY: Prentice Hall, 4th ed. edition, 2010.

[13] E.M. Stein and R. Shakarchi.Fourier analysis, volume 1 ofPrinceton Lectures in Analysis. Princeton University Press, Princeton, NJ, 2003. An introduction.

[14] E.M. Stein and R. Shakarchi.Real analysis. Princeton Lectures in Analysis, III. Princeton University Press, Princeton, NJ, 2005. Measure theory, integration, and Hilbert spaces.

[15] E.M. Stein and R. Shakarchi.Functional analysis, volume 4 ofPrinceton Lectures in Analysis. Princeton University Press, Princeton, NJ, 2011. Introduction to further topics in analysis.

[16] D. Werner.Einf¨uhrung in die h¨ohere Analysis. Springer-Lehrbuch. [Springer Textbook]. Springer-Verlag, Berlin, corrected edition, 2009.

[17] D. Williams.Probability with martingales. Cambridge Mathematical Textbooks. Cambridge University Press, Cambridge, 1991.

[18] A. Zygmund. The role of Fourier series in the development of analysis. InProceedings of the American Academy Workshop on the Evolution of Modern Mathematics (Boston, Mass., 1974), volume 2, pages 591–594, 1975.

[19] A. Zygmund.Trigonometric series. Vol. I, II. Cambridge Mathematical Library. Cambridge University Press, Cambridge, third edition, 2002. With a foreword by Robert A. Fefferman.